14. Given an arithmetic sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{3}=7, a_{5}+a_{7}=26 \text {, }
$$
The sum of the first $n$ terms of $\left\{a_{n}\right\}$ is $S_{n}$.
(1) Find $a_{n}$ and $S_{n}$;
(2) Let $b_{n}=\frac{1}{a_{n}^{2}-1}\left(n \in \mathbf{Z}_{+}\right)$, find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$, $T_{n}$.
14. (1) Let the common difference of the arithmetic sequence $\left\{a_{n}\right\}$ be $d$.

From the given information, we have
$$
\left\{\begin{array}{l}
a_{1}+2 d=7, \\
2 a_{1}+10 d=26
\end{array} \Rightarrow a_{1}=3, d=2\right. \text {. }
$$

Thus, $a_{n}=3+2(n-1)=2 n+1$,
$$
S_{n}=3 n+\frac{n(n-1)}{2} \times 2=n^{2}+2 n \text {. }
$$
(2) From (1), we know $a_{n}=2 n+1$. Therefore,
$$
\begin{array}{l}
b_{n}=\frac{1}{a_{n}^{2}-1}=\frac{1}{(2 n+1)^{2}-1} \\
=\frac{1}{4} \cdot \frac{1}{n(n+1)}=\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+1}\right) .
\end{array}
$$

Thus, $T_{n}=\frac{1}{4}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n}-\frac{1}{n+1}\right)$
$$
=\frac{1}{4}\left(1-\frac{1}{n+1}\right)=\frac{n}{4(n+1)} \text {, }
$$

which means the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $T_{n}=\frac{n}{4(n+1)}$.