(1) 求数列 $\left\{a_n\right\}$ 的通项公式;
(2) 求证: $\sin a_n-a_n < 0$;
(3) 证明: $\left(1+\sin \frac{1}{a_1}\right)\left(1+\sin \frac{1}{a_2}\right)\left(1+\sin \frac{1}{a_3}\right) \cdots\left(1+\sin \frac{1}{a_n}\right) < \mathrm{e}^2$.

(1) $\because 3 S_n=(n+2) a_n$ (1)

$\therefore 3 a_n=(n+2) a_n-(n+1) a_{n-1},(n-1) a_n=(n+1) a_{n-1}$, 即 $\frac{a_n}{n+1}=\frac{a_{n-1}}{n-1}$, 变形为 $\frac{a_n}{n(n+1)}=\frac{a_{n-1}}{(n-1) n}=\cdots=\frac{a_1}{1 \times 2}=\frac{1}{2} \therefore a_n=\frac{1}{2} n(n+1)(n \geq 2) \quad n=1$ 时也适合. $\therefore a_n=\frac{1}{2} n(n+1)\left(\mathrm{n} \in N^*\right)$
（2）构造函数 $F(x)=\sin x-x(x > 0), F^{\prime}(x)=\cos x-1 \leq 0, \therefore F(x)$ 在 $(0,+\infty)$ 上递减, $\therefore F(x) < F(0)=0, \therefore x > 0$ 时 $\sin x < x$. 令 $x=a_n$ 则有 $\sin a_n-a_n < 0$
(3) $\because \frac{1}{a_n} \in(0,1], \therefore \sin \frac{1}{a_n} > 0$, 原不等式等价于证明:

$$\ln \left(1+\sin \frac{1}{a_1}\right)+\ln \left(1+\sin \frac{1}{a_2}\right)+\cdots+\ln \left(1+\sin \frac{1}{a_{n-1}}\right)+\ln \left(1+\sin \frac{1}{a_n}\right) < 2,$$
$\because \ln (1+x) < x(x > 0)$, (证明略) $\quad \therefore \ln \left(1+\sin \frac{1}{a_n}\right) < \sin \frac{1}{a_n} < \frac{1}{a_n}=\frac{2}{n(n+1)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right)$ 令 $n=1,2,3 \cdots n$, 然后累加得
$$\ln \left(1+\sin \frac{1}{a_1}\right)+\ln \left(1+\sin \frac{1}{a_2}\right)+\cdots+\ln \left(1+\sin \frac{1}{a_{n-1}}\right)+\ln \left(1+\sin \frac{1}{a_n}\right) < 2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n}-\frac{1}{n+1}\right)$$
$=2\left(1-\frac{1}{n+1}\right) < 2$. 原不等式得证。