2. We prove the general case
$$\sqrt{2 n+1} \leqslant a_{n} \leqslant \sqrt{3 n+2} \text { (for all } n \geqslant 0 \text { ) }$$

Taking $n=1371, \sqrt{2 n+1}=\sqrt{2743} \approx 52.37, \sqrt{3 n+2}=\sqrt{4115} \approx 64.148$, then we have $520$. From equation (2), we know the sequence $\left\{a_{n}\right\}$ is strictly monotonically increasing, and $\frac{1}{\left(a_{n-1}\right)^{2}} \leqslant 1$ (for all $n \geqslant 1$). Also, since $a_{n}^{2}=a_{n-1}^{2}+\frac{1}{a_{n-1}^{2}}+2$, we get
a_{n-1}^{2}+2\sqrt{a_{n}^{2}+2} \geqslant \sqrt{2 n+1+2}=\sqrt{2(n+1)+2} \end{array}

Thus, the conclusion holds for $n+1$. By induction, equation (1) is true.