Queenly Queens

On any square the white queen is placed on, the black queen cannot be placed on the same rank or file. This leaves $64-8-7=49$ squares for the black queen to be placed.

However, we must also subtract the number of diagonal squares the white queen is attacking.

There are 2 such diagonals, from bottom left to top right or from top left to bottom right. They are symmetric, so we only have to consider one case and then multiply by 2. WLOG let us consider the bottom left to top right case.

When the white queen is on the A7-B8 diagonal, she attacks 1 square but there are 2 possible squares for her own placement. Thus we subtract $1*2$ .

Doing this process over all 13 such diagonals, we get $1*2+2*3+\ldots+6*7+7*8+6*7+5*6+\ldots+1*2$ $=2\sum_{n=1}^6 n(n+1)+7*8$ $2(\frac{6*7*13}{6}+\frac{6*7}{2})+56$ $=280$

Multiplying by 2 for the other type of diagonal and subtracting this from our previous total number of squares, $64*49-2*280=\boxed{2576}$