a) Let $|W_1B_1| = a_1$, $|W_2B_1| = a_2$, $|W_1B_2| = b_1$, and $|W_2B_2| = b_2$.

We are given the following conditions:
1. $a_1 + a_2 = 4$
2. $b_1 + b_2 = 8$
3. $b_1 + a_1 = 9$
4. $b_2 + a_2 = 3$

From these conditions, we can derive the following steps:

1. Subtract equation (4) from equation (3):
$$(b_1 + a_1) - (b_2 + a_2) = 9 - 3$$
$$b_1 + a_1 - b_2 - a_2 = 6$$

2. Substitute $a_1 + a_2 = 4$ into the equation:
$$b_1 - b_2 = 6$$

3. From equation (3), solve for $b_1$:
$$b_1 = 9 - a_1$$

4. From equation (4), solve for $b_2$:
$$b_2 = 3 - a_2$$

5. Substitute $b_1 = 9 - a_1$ and $b_2 = 3 - a_2$ into $b_1 - b_2 = 6$:
$$(9 - a_1) - (3 - a_2) = 6$$
$$9 - a_1 - 3 + a_2 = 6$$
$$6 - a_1 + a_2 = 6$$
$$-a_1 + a_2 = 0$$
$$a_1 = a_2$$

6. Since $a_1 + a_2 = 4$ and $a_1 = a_2$:
$$2a_1 = 4$$
$$a_1 = 2$$
$$a_2 = 2$$

7. Substitute $a_1 = 2$ into $b_1 = 9 - a_1$:
$$b_1 = 9 - 2 = 7$$

8. Substitute $a_2 = 2$ into $b_2 = 3 - a_2$:
$$b_2 = 3 - 2 = 1$$

Since $b_1 = 7$ is greater than the other distances, the extremes are $W_1$ (first white cat) and $B_2$ (second black cat).

The final answer is $\boxed{ W_1 }$ and $B_2$.