Answer: 608 Consider first placing the white rooks. They will occupy either 3 columns and 1 row, 3 columns and 2 rows, 3 columns and 3 rows, 2 rows and 2 columns, 2 columns and 3 rows, or 1 column and 3 rows. First note that placing the black rooks is impossible in the second, third, and fifth cases. The first case can happen in $4 \cdot 4$ ways, and each leads to a unique way to place the black rooks. In the fourth case, we can choose the row with 2 rooks in 4 ways, place the rooks in $\binom{4}{2}$ ways, choose the column of the other rook in 2 ways, and place it in 3 ways, for a total of $4 \cdot\binom{4}{2} \cdot 2 \cdot 3=144$ ways to place the white rooks in this configuration; the black rooks can then be placed in any of 4 possible locations, and there are 4 ways to do this, leading to 576 possibilities in this case. Finally, the sixth case is analogous to the first, giving 16 possibilities. Summing, we find $16+576+16=608$ total placements.
3. [5] Joe B. now clears the board. How many ways can he place 3 white rooks and 3 black rooks on the board so that no two rooks of opposite color can attack each other?