Here are the solutions, as written by Pascal Wassong, harry Nelson, Jim Boyce, and Andrej Jakobcic

1)
Raymond Smullyan, Chess problems with Sherlock Holmes, 1979
White : Kc8; Bg1; Ph2 : 3
Black : Ka8 : 1
Last move ? White to play.

Ka7xNa8

2)
Jean-Claude Dumont & Jean-Claude Gandy, Europe Echecs, 1977
White : Ka4; Rb4; Bg1, g6; Pc4, c6, h2 : 7
Black : Ka6; Pa7, b7 : 3
Mate in 2 moves. Last moves ?

White plays forward 1.Bg6-f5 and mates by 2.Bf5-c8# or 2.c4xb5#
depending on black move.
In the position of the diagram, black played the last move. It can only
have been Kb6(x)a6, eventually capturing on a6. The double check by the
wRb4 and wBg1 on b6 is explained by the en passant capture b5xc6 e.p.
The previous black move is therefore c7-c5. Then the check by wBg1 can
only be explained by Nc5(x)a6. So we see that the bK captured a wNa6. To
avoid retro-pat by the black, this wN must have captured a bNa6 to allow
bNb8(x)a6. Here ends the sequence of moves that we can deduce. So the
last moves were :
White : Ka4; Rb4; Bg1, g6; Nc5; Pb5, c4, h2 : 8 (and eventually
something on a6)
Black : Kb6; Nb8; Pa7, b7, c7 : 5
1. ......       Nb8(x)a6
2. Nc5xa6+      c7-c5
3.b5xc6 e.p.++  Kb6xa6 diagram.

3)
Gideon Husserl, Israel Ring Tourney, 1966 - 1971
White : Kc8, d6; Qc6; Rd8, f6; Ba8; Ne8; Pa7, b7 : 9
Black : 0
Color the pieces. Last move ?

The Qc6 and the Rd8 give to check to both Kings. This can only be
explained by these men being white and the last move being c7xd8=R++
with bKc8 and wKd6. Pb7, Ne8 and Rf6 must not give check. So we have
bPb7, wNe8 and wRf6. Because of bPb7, the position of the Ba8 can only
be explained by and under-promotion : wBa8 and wPa7.
The black piece captured at d8 cannot be a Rook or a Queen because they
make an unexplained check to the wK. If it is a bB, black doesn't have a
previous move. So it must be a bN which played bNe6(x)d8. The last move
is completely determined : c7xNd8=R++.

4)
Tibor Orban, Die Schwalbe, 1976
White : Ke1; Qd1; Ra1, h1; Bc1; Nb1, g1; Pa2, b2, c2, d2, e4, f2, g2, h2
: 15
Black : Ke8; Qd8; Ra8, h8; Bc8, f8, Nb8, g8; Pa7, b7, c6, e6, f7, g7, h7
: 15
Proof game in exactly 4 moves.

There are easy solutions in 3 moves, e.g. : 1.e4 c6 2.Bc4 e6 3.Bxe6 dxe6
diagram.
But in 4 moves the solution is unique : 1.e4 e6 2.Bb5 Ke7! 3.Bxd7 c6
4.Be8 Kxe8 diagram.

5)
The intention is (backward) :

1.Bf7-e6+ g4xPf3
2.Bh6-g5 g5-g4
3.Qh5-g6 g6-g5
4.Rg5-f5 f5-f4
5.Rd6-f6 f6-f5
6.Fe6-f7 f7-f6
7.Nf6-g8 Kg8-f8
8.Ng4-f6+ etc...

The 3rd Knight comes from the promotion on a8 after 2 captures axNb and
bxRa7-a8=N.
The wPg has been captured on his column.
The 15 last single moves are completely determined.

6)
Black captured Ph7xg6xf5xe4xd3xc2. After that white could play d2-d3 and
then the wBc1 could get captured by the bPa on b6 or b4. After that the
wPa promoted on a8 to replace the wBe2 or the wRh1 already captured by
the bP.

We will suppose that white may castle. So the wRh1 never moved. So the
wPa promoted to the wB currently at e2. I will now show that this
implies that the bK must have moved.

When black played h7xg6, the bPf7 must already have been on f6 so that
the move Px? isn't ambiguous. When black played g6xf5, the bPe7 is still
at e7 for the same reason.

The U-Chess rules make that the first move of each side can only be
played by the d or e pawn. The wPd played after the bPh reached the c2
square. The bPe7 moved after bPh7 reached f5. So we see that this game
begun by 1.e3 d6.

What is important here is the first black move. We deduce that when
black played e7-e6 to free the bBf8, the bPf was at f6 and the bPd at
d6. This Bishop passed by the squares e7, d8 and over b6 to reach the
his destination of the diagram c5.

At the moment of the first move of the promoted white bishop Ba8-b7, the
bPb6 is already there. This means that the bBc5 is also at his final
position and so we know that black already played c7-c6. Because the
wPc4 and wPd3 must have reached there final square to allow the wBc1 to
be captured by the bPa, the only way for the for the promoted wBa8 to
reach e2 is through b7, c8, d7, e8 at least, showing that the bK must
have evacuated his square.

So we showed that if white may castle, black cannot. If we suppose that
black can castle, this prouve us that the promoted piece on a8 is the
wRh1 and so white cannot castle.

This is an interesting problem like they were composed in those years.

7)
Dr. Luigi Ceriani, problem, 1952
White : Kg2; Qg1; Rh2; Bc1, f8, h1; Nd1; Pb2, c2, d2, e3, f2, g3, h3 :
14
Black : Ke2; Pb7, d5, e6, e7, g7, h5 : 7
What was the first move of the black king ?

The wPa2 captured a2xb3xc4xd5xe6xf7-f8=B. The bBf8 was captured at home.
The bPd7 captured on e6 after the wPa2 arrived at f7. The bPd5 comes
from c7 capturing the second missing white piece.
To open the South cage, the only way to put a piece on f1 and to retract
Ke1-e2. This piece can only have been a bR captured on e1 by a wN. This
wN was captured by the bPc7.
The presence of the bR behind the wall of the wPawns is explained by the
cross-captures h2xg3 and g2xh3. We now know where all the pieces have
been captured.
The last moves are : 1. ... c6xNd5 2.Nb4-d5 c7-c6 3.Nd3-b4 h6-h5
4.Ne1-d3 h7-h6 5.Nd3xRe1 Rf1-e1 etc...
We can now deduce that bPf7 has been captured at f7 without moving. So
the wPa2 didn't capture the bRh8 (currently at e1), the bBc8 and the
bQd8, because they couldn't leave the 8th row before bPd7xe6.
At the moment of the move f7-f8=B the pieces must have been at : bBc8,
bQd8, bRe8 and bKh8.
The place of the BRe8 and bKh8 imposes as there first move : O-O !
Harry Nelson mentions that Black's a-pawn must promote in this problem. His shortest game for this position is 43 moves.  Is that minimal?

8)
Gianni Donati, Die Schwalbe, 1998
Dedicated to Danna
White : Ke1; Qd1; Ra1, h1; Bc1, f1; Ng1; Pa2, b2, c2, d2, f2, g2, h2 :
14
Black : Kg8; Qa4; Ra8, h8; Be2, g3; Nb8, g6; Pa7, b4, c7, d7, f7, g7, h7
: 15
Shortest Proof Game in 12 moves.

1.Nc3 b5 2.Nd5 Ba6 3.Nxe7 Nxe7 4.e4! Ng6 5.e5 Bd6 6.e6 Kf8 7.e7+ Kg8
8.e8=N! Bg3 9.Nd6 Qh4 10.Nf5 Qa4 11.Nd4 b4 12.Nde2 Bxe2 diagram.
The fact that after his third move white must play an odd number of
moves imposes him to promote his e pawn to a Knight and this N must come
back to its original square to be captured ! A very beautifull problem.

Jim Boyce's solutions to 5 and 6:
Hi,

The last several moves leading to the Scherer position were

ng4-f6+  kg8-f8
nf6-g8   pf7-f6
be6-f7   pf6-f5
rd6-f6   pf5-f4
rg4-f4   pg6-g5
qh5-g6   pg5-g4
bh6-g5   pg4xpf3
bf7-e6+

Reasoning:

1) Black still has all his pawns and six of eight pieces.  He is
missing a knight and a rook.  The rook spen all it time on a8, a7, or
b8.

2) White has promoted a pawn to a knight.  White is missing one unit,
which was captured on the f-file by the g-pawn.

3) Where are White's a- and b-pawns?  One is on b6.  The other was
either captured on the f-file or is the extra knight.  In order to get
to the f-file, it must have been promoted.  So one of the a- and
b-pawns has been promoted.  It took a path through b6, a7 (capturing
the black rook), and a8.  The other capture moved the a-pawn to the
b-file.

4) So the White f-pawn stayed on the f-file until it was captured by
the Black g-pawn.

5) Now we can start retracting moves.  The final move was a discovered
check.  The moving piece was the bishop, and not a pawn being promoted
to a knight.

6) The only move black could have made was to capture on f3, and to
captured unit was a pawn.

7) White's next several retractions must leave Black with a legal
retraction.  White retracts moves leaveing squares for the f-and
g-pawns to have moved from.

8) Finally, White retracts his two knight moves, the earliest of which
was a check.

-----

Now for the U-chess problem.

1) White has lost 6 units; Black has made six captures with pawns.
Five of those captures were made on white squares (g6, f5, e4, d3, and
c2).  White's dark-square bishop was captured by the a-pawn.

2) White's d-pawn did not move until the five white square captures
were done.  Then the d-pawn moved to d3.  By this time, the c-pawn has
gone to c4, and freed the dark-square bishop to go die on the b-file.

3) White has not captured anything.  So White's a-pawn was promoted
only after Black cleared the a-file.  So the promoted a-pawn is still
on the board.

4) Now for some U-chess stuff.  The Black f-pawn was already on f6
when the g6 capture happened.  The Black e-pawn was still on e7 when
the f5 capture happened.

5) Black's first move was p-q3.

6) Since the pawns were already on f6 and d6 when the pawn moved to
e6, the path for the Black dark-square bishop to get to c5 goes
through e7 and d8 and over c7 and b6.  So the c-pawn moved to c6
before the b-pawns moved to b6.

7) Which is White's promoted piece?
7a) If it is the rook, then White cannot castle.
7b) If it is the bishop, then Black cannot castle, because the path
from a8 to e2 must pass through e8.  Several squares were already
obstructed before the b-pawn moved to b6: c2, c4, c6, and e6.

8) It seems clear that a games can be constructed to show that White
can castle, or that Black can castle.  Which one depends on what the
a-pawn was promoted to.

Pascal Wassong
Europe Echecs, 1994
version
Dedicated to Babette
White : Kc5; Qa2; Rb1, b2; Be1, h8; Na6, d2; Pa4, b3, b4, c2, e3, f2 :
14
Black : Kc3; Qc1; Rd1, e2; Ba7; Na1; Pc6, c7, d7, e6, f3, g7 : 12
What where the last 56 single moves ?

I don't give all the explanations. This is a complicated problem, were
some tries are not so easy to eliminate. Here is the untangled position
:
White : Kc5; Qa4; Ra1, c1; Be1; Nb8, d2; Pa2, b3, b4, c2, e3, f2, h2 :
14
Black : Kb2; Qa5; Rd1, e2; Ba7, b5; Nb6; Pc6, c7, d7, e6, f7, g7 : 14
And now forward :
1. ... Kb2-c3
2.Qa4-a3 Qa5-a4
3.Rc1-b1 a6-a5
4.Qa3-c1 Qa4-a3
5.Nb8-a6 Qa3-b2
6.Na6-b8 a5-a4
7.Nb8-a6 a4-a3
8.Na6-b8 Bb5-c4
9.Nb8-a6 Ba7-b8
10.h2-h3 Nb6-a4+
11.b3xa4 Bc4-b3
and from now on the moves are unique. Before we could have added extra
moves by the wNa6 and the bBc4, so that the previous move could have
been Bd5-b3.
12.a2xb3 a3-a2
13.h3-h4 Qb2-a3
14.Rb1-b2 f7-f6
15.Ra1-b1 a2-a1=N
16.Rb2-a2 Qa3-b2
17.Ra2-a3 Qb2-a2
18.Rb1-b2 Qa2-b1
19.Rb2-a2 Qb1-b2
20.Qc1-b1 Qb2-c1
21.Ra2-b2 f6-f5
22.Qb1-a2 Qc1-b1
23.h4-h5 Qb1-c1
24.Rb2-b1 Qc1-b2
25.Rb1-c1 Qb2-b1
26.h5-h6 Qb1-b2
27.Qa2-b1 Qb2-a2
28.h6-h7 Qa2-b2
29.Ra3-a2 Qb2-a3
30.Ra2-b2 Qa3-a2
31.h7-h8=B Qa2-a3
32.Qb1-a2 f5-f4
33.Rb2-b1 Qa3-b2
34.Qa2-a3 Qb2-a2
35.Rb1-b2 Qa2-b1
36.Rb2-a2 Qb1-b2
37.Rc1-b1 Qb2-c1
38.Ra2-b2 f4-f3
39.Qa3-a2 Bb8-a7+ diagram