a) Firstly, observe that if we change the colour of one square 3 times, it goes back to its original colour. As the final configuration does not depend on the order of moves and placing a move on a square 3 times does not change the configuration, we can suppose that a move is placed on any square exactly 0, 1 or 2 times. This further implies that a white square must change its colour 2 (modulo 3) times and that a black square must be changed 1 (modulo 3) times.
Let us associate each $2 \times 2$ square with its top left unit square. Let's take a look at the first column. Its top square is white and is included in only one $2 \times 2$ square, so that square has to be placed 2 times. The second square in this column is also white and is included in its $2 \times 2$ square as well as in the one associated with the first unit square in the column. This square has already been turned black by taking the $2 \times 2$ square of the first unit square in this column twice, so its own $2 \times 2$ square has to be taken 0 times.
We use similar arguments to show that next four $2 \times 2$ squares have to be called 1,1,1 and 0 times. After that we have the white square included only in its own $2 \times 2$ square and in the one above it, which has been taken 0 times, so we have the same situation we had in the beginning. Therefore, the next one should be used 2 times and the next is white, so 0 times. However, this is the only $2 \times 2$ that contains the last black square, therefore, this cannot change color, a contradiction.

b) Working as in a), we can fill the board with the numbers 0, 1, 2 in each square, showing how many times we should use the corresponding $2 \times 2$ square.