14. Given an arithmetic sequence $\left\{a_{n}\right\}$ satisfies: $a_{3}=7, a_{5}+a_{7}=26$, and the sum of the first $n$ terms of $\left\{a_{n}\right\}$ is $S_{n}$.
(1) Find $a_{n}$ and $S_{n}$;
(2) Let $b_{n}=\frac{1}{a_{n}^{2}-1}\left(n \in \mathbf{N}_{+}\right)$, find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$, $T_{n}$.
(1) Let the common difference of $\left\{a_{n}\right\}$ be $d$, then $\left\{\begin{array}{l}a_{3}=7, \\ a_{6}=13\end{array} \Rightarrow d=\frac{a_{6}-a_{3}}{6-3}=2 \Rightarrow a_{n}=2 n+1\right.$, so $S_{n}=n^{2}+2 n\left(n \in \mathbf{N}_{+}\right)$.
(2) $b_{n}=\frac{1}{(2 n+1)^{2}-1}=\frac{1}{4 n(n+1)}=\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+1}\right)$,

thus $T_{n}=\frac{1}{4}\left(1-\frac{1}{n+1}\right)=\frac{n}{4(n+1)}\left(n \in \mathbf{N}_{+}\right)$.