Example 1 (2005 Hope Cup Training Question) Given $a_{1}=\frac{1}{2}, a_{n-1}=\frac{a_{n}+3}{2 a_{n}-4}\left(n \geqslant 1, n \in \mathbf{N}^{*}\right)$, find the general formula for $a_{n}$.
Analysis: This problem is a recursive sequence of the type $a_{n+1}=\frac{\alpha_{1} a_{n}+\beta_{1}}{\alpha_{2} a_{n}+\beta_{2}}$, and the fixed point method can be used to solve it.
Solution: By solving the equation $x=\frac{x+3}{2 x-4}$, we get $x_{1}=-\frac{1}{2}, x_{2}=3$.
Therefore, $a_{n+1}+\frac{1}{2}=\frac{2\left(a_{n}+\frac{1}{2}\right)}{2 a_{n}-4} \cdots$ (1)
$$
a_{n+1}-3=\frac{-5\left(a_{n}-3\right)}{2 a_{n}-4} \cdots
$$
Dividing (1) by (2) gives
$$
\frac{a_{n+1}+\frac{1}{2}}{a_{n+1}-3}=-\frac{2}{5} \frac{a_{n}+\frac{1}{2}}{a_{n}-3}=\left(-\frac{2}{5}\right)^{n}\left(\frac{a_{1}+\frac{1}{2}}{a_{1}-3}\right) \text {. }
$$

Thus, $a_{n}=\frac{(-5)^{n}+3 \cdot 2^{n+1}}{2^{n+1}-2 \cdot(-5)^{n}}\left(n \in \mathbf{N}^{*}\right)$.
Evaluation: This is a typical problem of finding the general term formula of a sequence using the fixed point method. The general solution method is:
1) Find the fixed points $x_{1}, x_{2}$ by solving the characteristic equation;
2) List the sequences $b_{n}=a_{n}-x_{1}, c_{n}=a_{n}-x_{2}$;
3) Derive the relationship from $\frac{b_{n}}{c_{n}}$, and thus solve for $a_{n}$.