One, (Full marks 20 points) Given that positive integers $p, q$ are both prime numbers, and $7 p+q$ and $p q+11$ are also prime numbers. Calculate the value of $\left(p^{2}+q^{p}\right)\left(q^{2}+p^{q}\right)$.
Given that $p q+11$ is a prime number, we know that $p q+11$ must be odd. Therefore, $p q$ is even. So, at least one of $p, q$ must be even. But since $p, q$ are both primes, one of $p, q$ must be 2. If $p=q=2$, then $p q+11=15$ is not a prime number. Therefore, $p, q$ cannot both be 2. So, one and only one of $p, q$ is 2.

If $p=2$, then $14+q$ and $2 q+11$ are both primes. If $q$ leaves a remainder of 1 when divided by 3, then $q+14$ is divisible by 3. If $q$ leaves a remainder of 2 when divided by 3, then $2 q+11$ is divisible by 3, which contradicts the fact that $14+q$ and $2 q+11$ are primes. Therefore, $q=3$. Checking, we find that $p=2, q=3$ meets the requirements.

If $q=2$, then $7 p+2$ and $2 p+11$ are both primes. If $p$ leaves a remainder of 1 when divided by 3, then $7 p+2$ is divisible by 3. If $p$ leaves a remainder of 2 when divided by 3, then $2 p+11$ is divisible by 3, which contradicts the fact that $7 p+2$ and $2 p+11$ are primes. Therefore, $p=3$. Checking, we find that $p=3, q=2$ also meets the requirements.
Whether $p=2, q=3$ or $p=3, q=2$,
we have $\left(p^{2}+q^{p}\right)\left(q^{2}+p^{q}\right)=\left(3^{2}+2^{3}\right)\left(2^{2}+3^{2}\right)$
$$
=17 \times 13=221 .
$$