Three. (20 points) Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}$. And for any positive integer $n$, we have
$$
a_{n+2}=2(n+2) a_{n+1}-(n+2)(n+1) a_{n}+\frac{n^{2}+3 n+1}{n+3} \text {. }
$$

Try to find the general term formula of the sequence $a_{n}$.
$$
\begin{array}{l}
\frac{a_{n+2}}{(n+2)!}=\frac{2 a_{n+1}}{(n+1)!}-\frac{a_{n}}{n!}+\frac{n^{2}+3 n+1}{(n+3)!} . \\
\therefore \frac{a_{n+2}}{(n+2)!}-\frac{1}{(n+3)!} \\
=2\left[\frac{a_{n+1}}{(n+1)!}-\frac{1}{(n+2)!}\right]-\left[\frac{a_{n}}{n!}-\frac{1}{(n+1)!}\right] .
\end{array}
$$

Therefore, $\left\{\frac{a_{n}}{n!}-\frac{1}{(n+1)!}\right\}$ is an arithmetic sequence.
$$
\because \frac{a_{1}}{1!}-\frac{1}{2!}=0, \frac{a_{2}}{2!}-\frac{1}{3!}=0 \text {, }
$$
$\therefore$ the terms of the sequence $\left\{\frac{a_{n}}{n!}-\frac{1}{(n+1)!}\right\}$ are all zero.
Hence $\frac{a_{n}}{n!}=\frac{1}{(n+1)!}$, i.e., $a_{n}=\frac{1}{n+1}$.