13. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
S_{n}=1-a_{n}\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$

where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $c_{n}=\frac{1}{1+a_{n}}+\frac{1}{1-a_{n+1}}$, and the sum of the first $n$ terms of the sequence $\left\{c_{n}\right\}$ be $P_{n}$, prove: $P_{n}>2 n-\frac{1}{5}$.
13. (1) From the problem, we have
$$
S_{n+1}=1-a_{n+1} \text {. }
$$

Subtracting the given equation from the above equation, we get
$$
a_{n+1}=-a_{n+1}+a_{n} \Rightarrow a_{n+1}=\frac{1}{2} a_{n}\left(n \in \mathbf{Z}_{+}\right)
$$

Since $a_{1}=\frac{1}{2}$, we have
$$
a_{n}=\frac{1}{2} \times\left(\frac{1}{2}\right)^{n-1}=\left(\frac{1}{2}\right)^{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(2) From the given information, we have
$$
\begin{array}{l}
P_{n}=\frac{1}{1+a_{1}}+\sum_{i=2}^{n}\left(\frac{1}{1-a_{i}}+\frac{1}{1+a_{i}}\right)+\frac{1}{1-a_{n+1}} \\
=\frac{2}{3}+\frac{1}{1-\left(\frac{1}{2}\right)^{n+1}}+2 \sum_{i=2}^{n} \frac{1}{1-a_{i}^{2}} \\
=\frac{2}{3}+\frac{2^{n+1}}{2^{n+1}-1}+2 \sum_{i=2}^{n} \frac{4^{i}}{4^{i}-1} \\
=\frac{2}{3}+\frac{2^{n+1}}{2^{n+1}-1}+2 \sum_{i=2}^{n}\left(1+\frac{1}{4^{i}-1}\right) .
\end{array}
$$

When $n=1$, $P_{1}=2>\frac{3}{2}$, the conclusion holds;
When $n \geqslant 2$,
$$
\begin{array}{l}
P_{n}>\frac{2}{3}+1+2(n-1)+2 \sum_{i=2}^{n} \frac{1}{4^{i}-1} \\
=2 n-\frac{1}{3}+\frac{2}{4^{2}-1}+2 \sum_{i=3}^{n} \frac{1}{4^{i}-1} \\
>2 n-\frac{1}{5}
\end{array}
$$