13. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=a, a_{n+1}=\frac{5 a_{n}-8}{a_{n}-1}\left(n \in N^{*}\right)$.
(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 $a$.
Proof: (1) Since $\frac{a_{n+1}-2}{a_{n+1}-4}=\frac{\frac{5 a_{n}-8}{a_{n}-1}-2}{\frac{5 a_{n}-8}{a_{n}-1}-4}=3 \cdot \frac{a_{n}-2}{a_{n}-4}$, the sequence $\left\{\frac{a_{n}-2}{a_{n}-4}\right\}$ forms a geometric sequence. Therefore, $\frac{a_{n}-2}{a_{n}-4}=\frac{a_{1}-2}{a_{1}-4} \cdot 3^{n-1}$, solving this gives $a_{n}=\frac{4 \cdot 3^{n-1}+2}{3^{n-1}+1}$. Thus, the general term formula of the sequence is $a_{n}=\frac{4 \cdot 3^{n-1}+2}{3^{n-1}+1}$.
(2) Solution 1: Since $a_{n}>3$ holds for any positive integer $n$, we have $a=a_{1}>3$. From (1), we know $\frac{a_{n}-2}{a_{n}-4}=\frac{a-2}{a-4} \cdot 3^{n-1}$. Let $b=\frac{a-2}{a-4}$, then $a_{n}=4+\frac{2}{b \cdot 3^{n-1}-1}$.
(1) When $3b \cdot 3^{n-1}-1>b \cdot 3^{n}-1$, then $\frac{2}{b \cdot 3^{n}-1}>\frac{2}{b \cdot 3^{n-1}-1}$, thus $a_{n+1}>a_{n}$, meaning the sequence $\left\{a_{n}\right\}$ is monotonically increasing, hence $a_{n}>3$, satisfying the condition;
(2) When $a=4$, by the condition, $a_{n}=4$, satisfying the condition;
(3) When $a>4$, $a-2>a-4>0$, then $b>1$,
Notice that $a_{n}=3+\frac{b \cdot 3^{n-1}+1}{b \cdot 3^{n-1}-1}$, and $b \cdot 3^{n-1}-1>0$, so $a_{n}>3$, satisfying the condition.
In summary, the range of the real number $a$ is $(3,+\infty)$.
Solution 2: Since $a_{n}>3$ holds for any positive integer $n$, we have $a=a_{1}>3$.
We will use mathematical induction to prove: When $a>3$, for any positive integer $n$, $a_{n}>3$.
When $n=1$, the conclusion is obviously true;
Assume that when $n=k(k \geq 1)$, the conclusion holds, i.e., $a_{k}>3$.
When $n=k+1$, note that $a_{k+1}=\frac{5 a_{k}-8}{a_{k}-1}=3+\frac{2 a_{k}-5}{a_{k}-1}$.
From $a_{k}>3$, we know $2 a_{k}-5>0, a_{k}-1>0$, thus $a_{k+1}>3$.
Therefore, the conclusion also holds for $n=k+1$.
According to the principle of induction, for any positive integer $n$, $a_{n}>3$.
In summary, the range of the real number $a$ is $(3,+\infty)$.