3.616 ** Let $a_{1}=1, a_{2}=3$, for all positive integers $n$ we have
$$
a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n} .
$$

Find all values of $a_{n}$ that are divisible by 11.
From the given, we have $a_{n+2}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)$. Let $b_{n+1}=a_{n+1}-a_{n}, n \geqslant 1$. Thus, $b_{n+2}-(n+2) b_{n+1}$. Therefore,
$$
\begin{aligned}
b_{n} & -n b_{n-1}=n(n-1) b_{n-2}=\cdots=n(n-1) \cdots 3 b_{2}=n!(n \geqslant 2), \\
a_{n} & =\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right)+a_{1} \\
& =b_{n}+b_{n-1}+\cdots + b_{2}+1=1!+2!+\cdots+n!.
\end{aligned}
$$

When $n \leqslant 10$, by calculation, we find that only $a_{4}=33=11 \times 3$, $a_{8}=46233=11 \times 4203$, and $a_{10}=4037913=11 \times 367083$ are divisible by 11.

When $n \geqslant 11$, since $11!+\cdots+n!$ is divisible by 11, $a_{n}=a_{10}+(11!+\cdots+n!)$ is also divisible by 11.
In summary, the terms divisible by 11 are $a_{4}, a_{8}$, and $a_{n}(n \geqslant 10)$.