(12) Let the sequence $\left\{a_{n}\right\}$ satisfy: $a_{1}=1, a_{2}=2, \frac{a_{n+2}}{a_{n}}=\frac{a_{n+1}^{2}+1}{a_{n}^{2}+1}(n \geqslant 1)$.
(1) Find the recursive relation between $a_{n+1}$ and $a_{n}$, i.e., $a_{n+1}=f\left(a_{n}\right)$;
(2) Prove: $63<a_{2008}<78$.
12 (1) Given $a_{1}=1, a_{2}=2, \frac{a_{n+2}}{a_{n}}=\frac{a_{n+1}^{2}+1}{a_{n}^{2}+1}(n \geqslant 1)$, it is easy to see that for all $n \geqslant 1, a_{n} \neq 0$. When $n \geqslant 1$, from
$$
\frac{a_{n+2}}{a_{n}}=\frac{a_{n+1}^{2}+1}{a_{n}^{2}+1}
$$

we can derive
$$
\frac{a_{n+2} a_{n+1}}{a_{n+1}^{2}+1}=\frac{a_{n+1} a_{n}}{a_{n}^{2}+1}
$$

thus
$$
\frac{a_{n+2}}{a_{n+1}+\frac{1}{a_{n+1}}}=\frac{a_{n+1}}{a_{n}+\frac{1}{a_{n}}}.
$$

By successively applying the above relation, we get
$$
\frac{a_{n+1}}{a_{n}+\frac{1}{a_{n}}}=\frac{a_{n}}{a_{n-1}+\frac{1}{a_{n-1}}}=\frac{a_{n-1}}{a_{n-2}+\frac{1}{a_{n-2}}}=\cdots=\frac{a_{2}}{a_{1}+\frac{1}{a_{1}}}=\frac{2}{1+\frac{1}{1}}=1,
$$

thus $a_{n+1}=a_{n}+\frac{1}{a_{n}}$.
(2) Clearly, from $a_{1}=1$ and $a_{n+1}=a_{n}+\frac{1}{a_{n}}(n \geqslant 1)$, we know: for all $n \geqslant 1$, $a_{n} \geqslant 1$ holds, hence $06022,
$$

so $63^{2}<a_{2008}^{2}<78^{2}$, which means $63<a_{2008}<78$.