13. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=a, a_{n+1}=\frac{5 a_{n}-8}{a_{n}-1}\left(n \in \mathbf{N}_{+}\right)$.
(1) If $a=3$, prove that the sequence $\left\{\frac{a_{n}-2}{a_{n}-4}\right\}$ 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 $a$.
(1) $\frac{a_{n+1}-2}{a_{n+1}-4}=\frac{3 a_{n}-6}{a_{n}-4}=3 \cdot \frac{a_{n}-2}{a_{n}-4}$, and $\frac{a_{1}-2}{a_{1}-4}=-1 \Rightarrow \frac{a_{n}-2}{a_{n}-4}=-3^{n-1}$,

so $\frac{2}{a_{n}-4}=-3^{n-1}-1 \Rightarrow a_{n}-4=-\frac{2}{3^{n-1}+1} \Rightarrow a_{n}=4-\frac{2}{3^{n-1}+1}\left(n \in \mathbf{N}_{+}\right)$.
(2) $\frac{a_{1}-2}{a_{1}-4}=\frac{a-2}{a-4} \Rightarrow \frac{a_{n}-2}{a_{n}-4}=\frac{a-2}{a-4} \cdot 3^{n-1} \Rightarrow \frac{2}{a_{n}-4}=\frac{a-2}{a-4} \cdot 3^{n-1}-1$
$\Rightarrow a_{n}=4+\frac{2}{\frac{a-2}{a-4} \cdot 3^{n-1}-1}>3 \Rightarrow \frac{2}{\frac{a-2}{a-4} \cdot 3^{n-1}-1}>-1$.
When $\frac{a-2}{a-4} \cdot 3^{n-1}-1>0$, $\left\{\begin{array}{l}a>4, \\ \frac{a-2}{a-4} \cdot 3^{n-1}>1\end{array} \Rightarrow\left\{\begin{array}{l}a>4, \\ 3^{n-1}>\frac{a-4}{a-2},\end{array} \Rightarrow a>4\right.\right.$;
When $\frac{a-2}{a-4} \cdot 3^{n-1}-1<0$, $\left\{\begin{array}{l}2<a<4, \\ 3^{n-1}<\frac{4-a}{a-2}\end{array} \Rightarrow\left\{\begin{array}{l}2<a<4, \\ 3^{n-1}<\frac{4-a}{a-2}\end{array} \Rightarrow 3<a<4\right.\right.$; Clearly, when $a=4$, the conclusion also holds.
In summary, the range of real number $a$ is $(3,+\infty)$.