Bobby can mate Garry in four moves:
1. d4 d5 2. Qd3 Qd6 3. Qh3 Qh6 4. Qxc8

First, we notice that since Black made the last move, either the king or the rook has been moved, consequently rendering Black unable to castle. Now White plays Qa1 and no matter what is Black’s next move, Qh8 gives check-mate.

The rules of FIDE state that if a player runs out of time, their opponent wins the game IF they have a path to victory. If there is no sufficient material, e.g. a King and a Knight against a King, then the game is declared a draw.
In this position, Black played Rxg6 which forces the moves:
… Rxg6+
Nxg6+ Rxg6+
Kxg6+ Qxg6+
Kxg6
This leaves White with a King and a Knight against Black’s King. Thus, White did not have a path to victory and the game was declared a draw.

The first move is 1. Qa8. If Black plays Qd4, then White responds with Re3#. Otherwise, White plays either Nc5# or Nd6#, so that Black cannot capture the Knight neither with the Rook or the Queen.

The sequence of moves is: 1. Kd6 … 2. Kd7 … 3. Qb5 … 4. Kc7 … 5. Qb7#

This game is a win for White.
1. c7 Rd6+
2. Kb5 Rd5+
3. Kb4 Rd4+
4. Kb3 Rd3+
5. Kc2! Rd4!
6. c8=R! Ra4
7. Kb3
Now Black will either lose the rook, or get mated in one. If White promoted a Queen instead of a Rook, then 6… Rc4+ would lead to 7. Qxc4, which is a stalemate.

The digits represent the minimum number of knight-moves needed to reach the respective cells, starting from the center.

The least number of moves required to switch the positions of the knights is 16. An example is shown below.
Next, we prove that it is impossible to switch the positions of the knights with less than 16 moves. Since the white knights occupy 2 black and 1 white squares, they need to end up on 2 white and 1 black squares, and each knight must make at least 2 moves in order to get to the opposite side, the total number of moves for the white knights should be an odd number, larger or equal to 2+2+2=6. The same applies to the total number of moves for the black knights. Therefore, the only possible way to get a total number of moves less than 16 is if both the white knights and the black knights move exactly 7 times.
We assume it is possible to switch the positions with 7+7=14 moves in total. Then, the white knight on A2 and one of the white knights on A1, A3 should make 2 moves each, and the third white knight should make 3 moves and land on a white square, either D1 or D3. Without loss of generality, we assume the knight on A1 makes 3 moves: A1-B3-C1-D3. Then, the knight on A2 should make 2 moves: A2-C3-D1, and the knight on A3 should make 2 moves: A3-B1-D2.
We make the same argument for the black knights. Since it is impossible that the white knight on A1 moves along the trajectory A1-B3-C1-D3 and also the black knight on D3 moves along the trajectory D3-C1-B3-A1, we conclude that the black knight on D1 should make 3 moves: D1-C3-B1-A3, the black knight on D2 should make 2 moves: D2-B3-A1, and the black knight on D3 should make 2 moves: D3-C1-A2.
This is only possible if:
the knight on A1 moves to C1 after the knight on D3 moves to A2
the knight on D3 moves to A2 after the knight on A2 moves to C3
the knight on A2 moves to C3 after the knight on D1 moves to B1
the knight on D1 moves to B1 after the knight on A3 moves to D2
the knight on A3 moves to D2 after the knight on D2 moves to B3
the knight on D2 moves to B3 after the knight on A1 moves to C1
We get a contradiction which means that the least number of moves is 16.

If we put 32 knights on all black squares, then no two pieces will attack each other. Now let’s see that if we have more than 32 pieces, then there will be two which attack each other. Split the chessboard in 8 rectangular sectors of size 2×4. It is not hard to see that if we have more than 4 pieces in the same 2×4 sector, then 2 of them will attack each other. Therefore we can place at most 4 × 8 = 32 pieces on the chessboard.

First, White plays f3 and threatens mate with Qxe2. Indeed, blocking with the black rook on d4 will not help, because it will become pinned, which means that the rook on d6 will become unpinned, which will make the bishop on b6 pinned, and that will unpin the knight on c7, resulting in a mate. Below are listed all variations of the game.
1. … Rd5 2. Qxe2#
2. … Bxa5 2. Kc8#
3. … Bxc7 2. Nxc7#
4. … Bxe8 2. Kxe8#
5. … Qxe7+ 2. Kxe7#
6. … Rd2 2. Bxd2#
7. … Rxd6+ 2. Qxd6#