16. Given the function $f(x)=\left(x^{2}-4\right)^{-\frac{1}{2}}(x>2), f^{-1}(x)$ is the inverse function of $f(x)$.
(1) If the sequence $\left\{a_{n}\right\}, a_{1}=1, \frac{1}{a_{n+1}}=f^{-1}\left(a_{n}\right)\left(n \in \mathbf{N}_{+}\right)$, find $a_{n}$;
(2) Let $S_{n}=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}, b_{n}=S_{2 n+1}-S_{n}$, does there exist a largest positive integer $p$, such that for all $n \in \mathbf{N}_{+}$, $b_{n}<\frac{1}{p}$ holds? Please provide an explanation or proof.
16. (1) From $\frac{1}{a_{n+1}}=f^{-1}\left(a_{n}\right)$, we get $a_{n}=f\left(\frac{1}{a_{n+1}}\right)=\left[\left(\frac{1}{a_{n+1}}\right)^{2}-4\right]^{-\frac{1}{2}}$, hence $\left(\frac{1}{a_{n}}\right)^{2}=\left(\frac{1}{a_{n+1}}\right)^{2}-4$, which means $\left(\frac{1}{a_{n+1}}\right)^{2}-\left(\frac{1}{a_{n}}\right)^{2}=4$. Therefore, $\left\{\left(\frac{1}{a_{n}}\right)^{2}\right\}$ is an arithmetic sequence with a common difference of 4 and the first term is 1, so $\left(\frac{1}{a_{n}}\right)^{2}=4 n-3$. Thus, $a_{n}=\frac{1}{\sqrt{4 n-3}}$.
(2) $b_{n+1}-b_{n}=\left(S_{2 n+3}-S_{n+1}\right)-\left(S_{2 n+1}-S_{n}\right)=\left(S_{2 n+3}-S_{2 n+1}\right)-\left(S_{n+1}-S_{n}\right)=a_{2 n+3}^{2}+$ $a_{2 n+2}^{2}-a_{n+1}^{2}=\frac{1}{8 n+9}+\frac{1}{8 n+5}-\frac{1}{4 n+1}=-\frac{40 n+31}{(8 n+9)(8 n+5)(4 n+1)}<0$, so $\left\{b_{n}\right\}$ is a decreasing sequence. Therefore, to make $b_{n}<\frac{1}{p}$ for all $n \in \mathbf{N}_{+}$, we only need $b_{1}=\frac{14}{45}<\frac{1}{p}$, solving for $p<\frac{45}{14}$. Thus, there exists a positive integer $p=3$ that satisfies the requirement.