13. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=a, a_{n+1}=\frac{5 a_{n}-8}{a_{n}-1}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(1) If $a=3$, prove that $\left\{\frac{a_{n}-2}{a_{n}-4}\right\}$ is a geometric sequence, and find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If for any positive integer $n$, $a_{n}>3$, find the range of the real number $a$.
Three, 13. (1) From the given information,
$$
\frac{a_{n+1}-2}{a_{n+1}-4}=\frac{\frac{5 a_{n}-8}{a_{n}-1}-2}{\frac{5 a_{n}-8}{a_{n}-1}-4}=3 \cdot \frac{a_{n}-2}{a_{n}-4}.
$$
Thus, $\left\{\frac{a_{n}-2}{a_{n}-4}\right\}$ is a geometric sequence.
$$
\begin{array}{l}
\text { Therefore, } \frac{a_{n}-2}{a_{n}-4}=\frac{a_{1}-2}{a_{1}-4} \cdot 3^{n-1}=-3^{n-1} \\
\Rightarrow a_{n}=\frac{4 \times 3^{n-1}+2}{3^{n-1}+1} .
\end{array}
$$
(2) Given that $a_{n}>3$ for any positive integer $n$, we know $a=a_{1}>3$.

We will prove by mathematical induction that when $a>3$, for any positive integer $n$, we have $a_{n}>3$.
When $n=1$, the conclusion holds.
Assume that for $n=k(k \geqslant 1)$, the conclusion holds.

When $n=k+1$, note that, $a_{k+1}=\frac{5 a_{k}-8}{a_{k}-1}=3+\frac{2 a_{k}-5}{a_{k}-1}$.
Since $a_{k}>3$, we have $2 a_{k}-5>0, a_{k}-1>0$.
Thus, $a_{k+1}>3$.
Therefore, the conclusion holds for $n=k+1$.
By the principle of induction, we know that for any positive integer, we have $a_{n}>3$.
In summary, the required real number $a \in(3,+\infty)$.