(2) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=2,(n+1) a_{n+1}=a_{n}+n\left(n \in \mathbf{N}^{*}\right)$, then $a_{n}=$ $\qquad$.
(2) $a_{n}=\frac{1}{n!}+1$ Hint: Since $(n+1) a_{n+1}=a_{n}+n$, so
$$
\begin{array}{l}
(n+1) a_{n+1}-(n+1)=a_{n}-1 \\
\Rightarrow(n+1)\left(a_{n+1}-1\right)=a_{n}-1 \\
\Rightarrow a_{n+1}-1=\frac{1}{n+1}\left(a_{n}-1\right)=\frac{1}{(n+1) n}\left(a_{n-1}-1\right)=\cdots \\
\quad=\frac{1}{(n+1) n(n-1) \cdots 2}\left(a_{1}-1\right)=\frac{1}{(n+1)!} \\
\Rightarrow a_{n+1}=\frac{1}{(n+1)!}+1 .
\end{array}
$$

Therefore, $a_{n}=\frac{1}{n!}+1$.