Example 6 Find the general term of the following recursive sequences:
$$
\begin{array}{l}
\text { (1) } a_{1}=\frac{1}{2}, a_{n+1}=\frac{a_{n}+3}{2 a_{n}-4}(n \geqslant 1) \text {; } \\
\text { (2) } a_{1}=5, a_{n+1}=\frac{a_{n}-4}{a_{n}-3}(n \geqslant 1) .
\end{array}
$$
(1) $a_{n+1}=f\left(a_{n}\right)=\frac{a_{n}+3}{2 a_{n}-4}$, where $f(x)=\frac{x+3}{2 x-4}$. Solving the equation $f(x)=x$, i.e., $\frac{x+3}{2 x-4}=x, 2 x^{2}-5 x-3=0$, we find that $f(x)$ has two fixed points $x_{1}=-\frac{1}{2}, x_{2}=3$, thus
$$
\begin{array}{l}
a_{n+1}-\left(-\frac{1}{2}\right)=\frac{a_{n}+3}{2 a_{n}-4}-\left(-\frac{1}{2}\right)=\frac{2\left(a_{n}+\frac{1}{2}\right)}{2 a_{n}-4}, \\
a_{n+1}-3=\frac{a_{n}+3}{2 a_{n}-4}-3=\frac{-5\left(a_{n}-3\right)}{2 a_{n}-4} .
\end{array}
$$

Since $a_{n} \neq 3$ (otherwise, from (2) we can deduce $a_{1}=3 \neq \frac{1}{2}$, which is a contradiction), dividing (1) by (2) gives
$$
\frac{a_{n+1}+\frac{1}{2}}{a_{n+1}-3}=\left(-\frac{2}{5}\right)\left(\frac{a_{n}+\frac{1}{2}}{a_{n+1}-3}\right) \text {. }
$$

Let $y_{n}=\frac{a_{n}+\frac{1}{2}}{a_{n}-3}$, then $y_{n}$ is a geometric sequence with the first term $y_{1}=\frac{a_{1}+\frac{1}{2}}{a_{1}-3}=-\frac{2}{5}$ and common ratio $-\frac{2}{5}$, thus $y_{n}=-\frac{2}{5}\left(-\frac{2}{5}\right)^{n-1}$, i.e., $\frac{a_{n}+\frac{1}{2}}{a_{n}-3}=\left(-\frac{2}{5}\right)^{n}$,
solving for $a_{n}$ gives $a_{n}=\frac{-5^{n}+(-1)^{n-1} 3 \cdot 2^{n+1}}{2 \cdot 5^{n}+(-1)^{n-1} \cdot 2^{n+1}}$.
(2) $a_{n+1}=f\left(a_{n}\right)=\frac{a_{n}-4}{a_{n}-3}$, where $f(x)=\frac{x-4}{x-3}$, solving the equation $f(x)=x$, i.e., $\frac{x-4}{x-3}=x$, $x^{2}-4 x+4=0$ we find that $f(x)$ has only one fixed point $x_{0}=2$, thus
$$
a_{n+1}-2=\frac{a_{n}-4}{a_{n}-3}-2=\frac{-a_{n}+2}{a_{n}-3} .
$$

Since $a_{n} \neq 2$ (otherwise, from (3) we can deduce $a_{1}=2 \neq 5$ which is a contradiction), we have
$$
\frac{1}{a_{n+1}-2}=\frac{-a_{n}+3}{a_{n}-2}=-1+\frac{1}{a_{n}-2},
$$

hence $\frac{1}{a_{n}-2}$ is an arithmetic sequence with the first term $\frac{1}{a_{1}-2}=\frac{1}{3}$ and common difference $d=-1$, so
$$
\frac{1}{a_{n}-2}=\frac{1}{3}+(n-1)(-1)=\frac{4}{3}-n=\frac{4-3 n}{3},
$$

solving for $a_{n}$ gives $a_{n}=\frac{3}{4-3 n}+2=\frac{6 n-11}{3 n-4}$.