If we don't care whether they attack each other, there are $64$ ways to place the White knight, and for each of those there are $63$ places for the black knight, so $64\times63=4032$ positions. From this we have to subtract the number of positions in which the knights attack each other.
Two knights (of different colors) attack each other if they are at opposite corners of a $3\times2$ or $2\times3$ rectangle, so we start by counting those rectangles.
A $3\times2$ rectangle is determined by its lower left corner square, which must be on one of the first $6$ files and one of the first $7$ ranks, so the number of such rectangles is $6\times7=42.$ The number of $2\times3$ rectangles is the same, so the total number of rectangles is $2\times42=84.$
For a pair of mutually attacking knights, the white knight must be at one of the $4$ corners of one of those $84$ rectangles, and the black knight at the opposite corner; so the number of positions for the two knights is $84\times4=336.$
The final answer is $\boxed{4032-336=3696}.$
Alternatively, let us define the degree of a square (in the knight's graph) to be the number of squares attacked by a knight on that square. By actual count there are $16$ squares of degree $8,$ $16$ of degree $6,$ $20$ of degree $4,$ $8$ of degree $3,$ and $4$ of degree $2;$ so the number of positions for two mutually attacking knights is $16\times8+16\times6+20\times4+8\times3+4\times2=336.$
More generally, the answer to the analogous problem on an $m\times n$ chessboard ($m,n\ge2$) is
$$mn(mn-1)-4[(m-2)(n-1)+(m-1)(n-2)].$$
"Knightwise" we have a circular graph with $8$ vertices, and on some of these vertices there is a knight. We have to compute the probability $p$ that two adjacent vertices are taken. There are $2^8=256$ equiprobable configurations, and we have to count the configurations containing at least one pair of taken adjacent vertices. These are the good configurations.
If there are two knights there are $8$ good configurations, if there are three knights there are $8\cdot 4+8=40$ good configurations, and if there are four knights all configurations except the two regular ones are good, making ${8\choose4}-2=68$ good configurations. All ${8\choose5}+\ldots+{8\choose8}=93$ configurations with $\geq5$ knights are good. Since there are $8+40+68+93=209$ good configurations we obtain
$$p={209\over256}\ .$$
Best Answer
First of all, here is the proof that you can put no more than $32$ non-attacking knights on a chessboard. You can pair up the squares of the board into $32$ pairs, where each pair is a knight's move apart, as shown below. Since you can place at most one knight in each pair, you can place at most $32$ knights.
Now, imagine placing a queen on some square in this same diagram, and then placing an X on every square attacked by the queen, as well as an X on the squares a knight's move away from the queen. No matter where you place the queen, at least $4$ of these pairs will have both of their squares X'ed out. Therefore, the number of knights you can place is reduced by at least four, so you cannot place more than $28$ knights.
For example, when the queen is placed at $\mathrm A1$, the lower left corner, then the pairs $\mathrm {A1-C2, A2-C1,B2-D1}$, and $\mathrm{A4-C3}$ are all X'ed out (the letter is the column, and the number is the row). It is not too hard to verify case by case that the queen always X's out at least $4$ of these pairs. In fact, as the queen moves closer to the center, she X's out more pairs.