8. A small bug crawls on the surface of a regular octahedron, moving from one vertex to one of the four adjacent vertices with equal probability each second. What is the probability that the bug returns to its initial position at the 8th second? $\qquad$ .
8. $\frac{43}{256}$.

As shown in Figure 3, after crawling $n$ steps, let the probability of the insect being at point $A$ be $p_{n}$, the probability of being at one of $B, C, D, E$ be $q_{n}$, and the probability of being at point $F$ be $r_{n}$.
$$
\begin{array}{l}
\text { Then } p_{n+1}=\frac{1}{4} q_{n}, \\
q_{n+1}=p_{n}+r_{n}+\frac{1}{2} q_{n}, \\
r_{n+1}=\frac{1}{4} q_{n} .
\end{array}
$$

Given $p_{1}=0, q_{1}=1, r_{1}=0$, by recursion we get $p_{8}=\frac{43}{256}$.