1 Let $p$ and $q$ both be prime numbers, and $7p + q$, $pq + 11$ are also prime numbers. Find the value of $\left(p^{2} + q^{p}\right)\left(q^{2} + p^{q}\right)$.
1. If $p$ and $q$ are both odd, then $7p + q$ is even, and it is not a prime number. Therefore, one of $p$ and $q$ must be even.

Case one: Suppose $p$ is even, then $p = 2$. In this case, since $7p + q$ is a prime number, $q$ must be an odd prime. If $q \neq 3$, then $q \equiv 1$ or $2 \pmod{3}$.

If $q \equiv 1 \pmod{3}$, then
$$7p + q = 14 + q \equiv 0 \pmod{3}$$

This is a contradiction;
If $q \equiv 2 \pmod{3}$, then
$$pq + 11 = 2q + 11 \equiv 4 + 11 \equiv 0 \pmod{3}$$

This is also a contradiction. Therefore, $q = 3$. In this case,
$$7p + q = 17, \quad pq + 11 = 17$$

Both are prime numbers, so
$$\left(p^2 + q^p\right)\left(q^2 + p^q\right) = \left(2^2 + 3^2\right)\left(3^2 + 2^3\right) = 221$$

Case two: Suppose $q$ is even, then $q = 2$. Similarly, we find that $p = 3$. In this case,
$$\left(p^2 + q^p\right)\left(q^2 + p^q\right) = \left(3^2 + 2^3\right)\left(2^2 + 3^2\right) = 221$$

In conclusion, the desired value is 221.