13. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $S_{n}=2 a_{n}+(-1)^{n}, n \geqslant 1$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove that for any integer $m>4$, $\frac{1}{a_{4}}+\frac{1}{a_{5}}+\cdots+\frac{1}{a_{m}}<\frac{7}{8}$.
13. Solution (1) When $n \geqslant 2$, we have $a_{n}=S_{n}-S_{n-1}=2\left(a_{n}-a_{n-1}\right)+2 \times(-1)^{n}$, so $a_{n}=2 a_{n-1}+2 \times (-1)^{n-1}$,
Solution 1: $a_{n}=2 a_{n-1}+2 \times(-1)^{n-1}, a_{n-1}=2 a_{n-2}+2 \times(-1)^{n-2}, \cdots a_{2}=2 a_{1}-2$.
Therefore, $a_{n}=2^{n-1} a_{1}+2^{n-1} \times(-1)+2^{n-2} \times(-1)^{2}+\cdots+2 \times(-1)^{n-1}$
$$
\begin{array}{l}
=2^{n-1}+(-1)^{n}\left[(-2)^{n-1}+(-2)^{n-2}+\cdots+(-2)\right] \\
=2^{n-1}-(-1)^{n} \frac{2\left[1-(-2)^{n-1}\right]}{3}=\frac{2}{3}\left[2^{n-2}+(-1)^{n-1}\right] .
\end{array}
$$

Upon verification, $a_{1}$ also satisfies the above formula, so $a_{n}=\frac{2}{3}\left[2^{n-2}+(-1)^{n-1}\right], n \geqslant 1$.
Solution 2: $\frac{a_{n}}{2^{n}}=\frac{a_{n-1}}{2^{n-1}}+\left(-\frac{1}{2}\right)^{n-1}$, the following uses the cumulative addition identity to prove it.
(2) Proof: From the general term formula, we get $a_{1}=2$.
When $n \geqslant 3$ and $n$ is odd, $\frac{1}{a_{n}}+\frac{1}{a_{n-1}}=\frac{3}{2}\left[\frac{1}{2^{n-2}+1}+\frac{1}{2^{n-1}-1}\right]$
$$
=\frac{3}{2} \times \frac{2^{n-1}+2^{n-2}}{2^{2 n-3}+2^{n-1}-2^{n-2}-1}
$$
When $n \geqslant 4$ and $m$ is even, $\frac{1}{a_{1}}+\frac{1}{a_{5}}+\cdots+\frac{1}{a_{m}}=\frac{1}{a_{4}}+\left(\frac{1}{a_{5}}+\frac{1}{a_{5}}\right)+\cdots+\left(\frac{1}{a_{m-1}}+\frac{1}{a_{m}}\right)$
When $n \geqslant 4$ and $m$ is odd, $\frac{1}{a_{4}}+\frac{1}{a_{5}}+\cdots+\frac{1}{a_{m}}$, we have $\frac{1}{a_{4}}+\frac{1}{a_{5}}+\cdots+\frac{1}{a_{m}}<\frac{7}{8}$.