3. (17th Japan Mathematical Olympiad) Find the tens digit of $11^{12^{13}}$ (where $11^{12^{3}}$ represents 11 to the power of $12^{13}$)
3. For any positive integer $n$, we have
$$
11^{n}=(10+1)^{n}=\sum_{k=0}^{n} C_{n}^{k} 10^{k} \equiv 10 n+1(\bmod 100) .
$$

Thus, the tens digit of $11^{n}$ is equal to the last digit of $n$.
Since the last digit of $12^{n}$ is $2,4,8,6,2,4, \cdots(n=1,2, \cdots)$, with a period of 4, the last digit of $12^{13}$ is 2.
Therefore, the tens digit of $11^{12^{13}}$ is 2.