II. (50 points) Find all prime pairs $(p, q)$ such that $p q \mid 5^{p}+5^{q}$.

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If $2 \mid p q$, without loss of generality, let $p=2$, then $2 q \mid 5^{2}+5^{q}$, so $q \mid 5^{q}+25$.
By Fermat's Little Theorem, $q \mid 5^{q}-5$, thus $q \mid 30$, i.e., $q=2,3,5$.
It is easy to verify that the prime pair $(2,2)$ does not meet the requirements, while $(2,3),(2,5)$ do meet the requirements.
If $p q$ is odd and $5 \mid p q$, without loss of generality, let $p=5$, then $5 q \mid 5^{5}+5^{q}$, so $q \mid 5^{q-1}+625$.
When $q=5$, the prime pair $(5,5)$ meets the requirements. When $q \neq 5$, by Fermat's Little Theorem, we have $q \mid 5^{q-1}-1$, hence $q \mid 626$. Since $q$ is an odd prime, and the only odd prime factor of 626 is 313, so $q=313$. After verification, the prime pair $(5,313)$ meets the requirements.
If $p, q$ are neither 2 nor 5, then $p q \mid 5^{p-1}+5^{q-1}$, so
$$
5^{p-1}+5^{q-1} \equiv 0(\bmod p) \text {. (1); }
$$

By Fermat's Little Theorem, we get $5^{p-1} \equiv 1(\bmod p)$, (2);
Thus, from (1) and (2), we have
$$
5^{q-1} \equiv-1(\bmod p) \text {. (3); }
$$

Let $p-1=2^{k}(2 r-1), q-1=2^{l}(2 s-1)$, where $k, l, r, s$ are positive integers.
If $k \leqslant l$, then from (2) and (3), it is easy to see that
$$
1=1^{2^{l-k}(2 x-1)} \equiv\left(5^{p-1}\right)^{2^{l-k}(2 s-1)}=5^{2^{l}(2 r-1)(2 s-1)}=\left(5^{q-1}\right)^{2 r-1} \equiv(-1)^{2 r-1} \equiv-1(\bmod p),
$$

which contradicts $p \neq 2$! Therefore, $k>l$.
Similarly, we have $k<l$, which is a contradiction! Thus, there are no $(p, q)$ that meet the requirements in this case.
In summary, all prime pairs $(p, q)$ that meet the requirements are
$$
(2,3),(3,2),(2,5),(5,2),(5,5),(5,313) \text { and }(313,5) \text {. }
$$