7.5. Given a 4x2017 board. Two players take turns placing rooks on it. The first player must place a rook so that it attacks an even number of already placed rooks, while the second player must place a rook so that it attacks an odd number of already placed rooks. The player who cannot make a move loses. Who can ensure their victory regardless of the opponent's actions? A rook attacks any distance along the same row or column, but it cannot jump over other pieces. For example, if three rooks are placed in the same row, the left rook attacks only the central one, and the central rook attacks both the right and left ones.
Answer: second

Solution: Let's mentally divide the $4 \times 2017$ board into two parts, each $2 \times 2017$. Then, let the second player mirror the first player's moves in the other part. For example, if the first player moves to the upper left cell of the first part, the second player moves to the other part, but also to the upper left cell. This way, the second player will always have a move, and thus will win since the game is finite. It remains to prove that the second player can indeed mirror the moves in this manner.

Notice that the second player, according to this strategy, always moves in the same column as the first player. Let's consider two cases:

1) There were no rooks in this column before the first player's move. Let's call the cell where the first player placed the rook A, and the cell where the second player wants to move B. The rook in A attacks an even number of other rooks, but the second player played symmetrically, so the rook in B also attacks the same number of rooks, plus the new rook in A, i.e., an odd number. Therefore, the move is possible.
2) There were already two rooks in this column. Notice that they are then placed one cell apart from each other. Again, let's call the cells A and B. Notice that the rows of these cells are the same, but from one of these cells, 2 rooks are attacked in this column, and from the other, 1 rook is attacked. That is, if the first player placed the rook in the cell from which an even number of rooks were attacked, then, firstly, he did not affect the other cell, and, secondly, from the other cell, an odd number of rooks are attacked, so the second player can now move there.

Criteria: only the strategy - 1 point. The strategy is correct, but one of the cases is omitted in the justification - 4 points.