(a) Suppose $m$ is odd, i.e., $m = 2k+1$ for some non-negative integer $k$. Let's color the board like a chessboard. In the first column, there will be $k+1$ squares of one color (say, white), and in the second column, there will be only $k$ squares of that color. However, each piece placed on a white square in the first column must also occupy a white square in the second column. This would mean that two pieces must occupy the same square in the second column, which is a contradiction. Therefore, $m$ must be even. By a similar argument, $n$ must also be even. Thus, the board must have dimensions $(2m, 2n)$ for some positive integers $m$ and $n$. (b) Let's color the $2m \times 2n$ board like a chessboard. Consider all the white squares on the perimeter of the board. There are a total of $2m + 2n - 2$ such white squares. Notice that each piece we use can occupy at most 2 of these perimeter white squares. Therefore, we need at least $(2m + 2n - 2)/2 = m+n-1$ pieces to cover the white perimeter squares. Similarly, we need at least $m+n-1$ pieces to cover the black perimeter squares. Thus, we need at least $2(m+n-1)$ pieces in total. It is easy to construct an example that achieves this bound (e.g., using diagonal strips).