11. Find all prime pairs $(p, q)$ such that $p q \mid\left(5^{p}-2^{p}\right)\left(5^{q}-2^{q}\right)$.
11. If $p \mid\left(5^{p}-2^{p}\right)$, then $p$ is coprime with 2 and 5. Using Fermat's Little Theorem, we know $5^{p}-2^{p} \equiv 5-2(\bmod p)$, hence $p \mid 3$. Therefore, $p=3$. In this case, $5^{p}-2^{p}=3^{2} \times 13$, so $q \mid\left(5^{q}-2^{q}\right)$ or $q \mid\left(3^{2} \times 13\right)$, which gives $q=3$ or 13. Similarly, discussing the case $q \mid\left(5^{q}-2^{q}\right)$, we get the solutions $(p, q)=(3,3),(3,13)$ or $(13,3)$.

Finally, suppose $p\left|\left(5^{q}-2^{q}\right), q\right|\left(5^{p}-2^{q}\right)$, and $p \neq 3, q \neq 3$. In this case, without loss of generality, assume $q < p$, then $(p, q-1)=1$. By Bézout's Theorem, there exist $a, b \in \mathbf{N}^{*}$ such that $a p-b(q-1)=1$. Combining $(5, q)=1,(2, q)=1$ and Fermat's Little Theorem, we know $5^{q^{-1}} \equiv 2^{q-1} \equiv 1(\bmod q)$. From $5^{p} \equiv 2^{p}(\bmod q)$, we also have $5^{a p} \equiv 2^{a p}(\bmod q)$, i.e.,
$$5^{1+b q-1)} \equiv 2^{1+b q-1)}(\bmod q)$$

This leads to $5 \equiv 2(\bmod q)$, i.e., $q \mid 3$, which is a contradiction.
Therefore, the prime pairs $(p, q)=(3,3),(3,13)$ or $(13,3)$ satisfy the conditions.