8. Given that $p$ and $q$ are both prime numbers, and $7p+q$, $2q+11$ are also prime numbers. Then $p^{q}+q^{p}=$ $\qquad$ .
8. 17 .

Since $7 p+q$ is a prime number, and $7 p+q>2$, $7 p+q$ must be an odd number.
Therefore, one of $p$ or $q$ must be even (which can only be 2).
Clearly, $q \neq 2$ (otherwise, $2 q+11=15$, which is not a prime number). Thus, $p=2$.
At this point, $14+q$ and $2 q+11$ are both prime numbers.
If $q=3 k+1\left(k \in \mathbf{N}_{+}\right)$, then
$$
14+q=3(k+5) \text {, }
$$

is not a prime number;
If $q=3 k+2\left(k \in \mathbf{N}_{+}\right)$, then
$$
2 q+11=3(2 k+5) \text {, }
$$

is not a prime number.
Therefore, $q=3 k\left(k \in \mathbf{N}_{+}\right)$, and the only solution is $q=3$.
Thus, $p^{q}+q^{p}=2^{3}+3^{2}=17$.