8 Given that $p$ and $q$ are both prime numbers, and $7p+q$ and $2q+11$ are also prime numbers. Then the value of $p^{q}+q^{p}$ is $\qquad$ .
817 Hint: Since $7 p+q$ is a prime number, and $7 p+q>2$, $7 p+q$ must be odd, so one of $p$ or $q$ must be even (and 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, both $14+q$ and $2 q+11$ are prime numbers.

If $q=3 k+1\left(k \in \mathbf{N}^{*}\right)$, then $14+q=3(k+5)$, which is not a prime number; if $q=3 k+2\left(k \in \mathbf{N}^{*}\right)$, then $2 q+11=3(2 k+5)$, which is not a prime number. Therefore, $q=3 k\left(k \in \mathbf{N}^{*}\right)$, and the only solution is $q=3$.
Thus, $p^{q}+q^{p}=2^{3}+3^{2}=17$.