II. (15 points) Given a sequence $\left\{a_{n}\right\}$ whose terms are all positive, the sum of its first $n$ terms is $S_{n}$, and for any $n \in \mathbf{Z}_{+}$, we have
$$
S_{n}^{2}-\left(n^{2}+n-1\right) S_{n}-\left(n^{2}+n\right)=0 \text {. }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{2}{(n+2) a_{n}}\left(n \in \mathbf{Z}_{+}\right)$, and the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $T_{n}$. Prove: $T_{n}<\frac{3}{4}\left(n \in \mathbf{Z}_{+}\right)$.
II. (1) From the problem, we know
$$
\left(S_{n}+1\right)\left(S_{n}-n^{2}-n\right)=0 \text {. }
$$

Since \( a_{n}>0 \), it follows that \( S_{n}>0 \left(n \in \mathbf{Z}_{+}\right) \).
Thus, \( S_{n}=n^{2}+n \).
When \( n \geqslant 2 \),
$$
\begin{array}{l}
a_{n}=S_{n}-S_{n-1} \\
=n^{2}+n-\left[(n-1)^{2}+(n-1)\right] \\
=2 n .
\end{array}
$$

Also, \( a_{1}=S_{1}=2 \) satisfies the above equation, therefore,
$$
a_{n}=2 n\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(2) From (1), we know \( a_{n}=2 n \).
$$
\begin{array}{l}
\text { Then } b_{n}=\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right) . \\
\text { Hence } T_{n}=b_{1}+b_{2}+b_{3}+\cdots+b_{n} \\
=\frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+2}\right)\right] \\
=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right) \\
<\frac{1}{2}\left(1+\frac{1}{2}\right)=\frac{3}{4} .
\end{array}
$$