Three. (20 points)
Given $a_{1}=a_{2}=1, a_{3}=2$, and $a_{n+1}=\frac{3+a_{n} \cdot a_{n-1}}{a_{n}-2} \quad(n \geqslant 3)$, find $a_{n}$.
Three, (20 points)
The original recurrence relation can be rewritten as:
$$
a_{n+1} \cdot a_{n-2}=3+a_{n} \cdot a_{n-1} .
$$

Substituting $n+1$ with $n$ in the above equation, we get
$$
a_{n} \cdot a_{n-3}=3+a_{n-1} \cdot a_{n-2}
$$

Subtracting the two equations, we obtain
$$
\begin{array}{c}
a_{n+1} \cdot a_{n-2} - a_{n} \cdot a_{n-3} = a_{n} \cdot a_{n-1} - a_{n-1} \cdot a_{n-2} \quad (n>3) \\
\frac{a_{n+1} + a_{n-1}}{a_{n}} = \frac{a_{n-1} + a_{n-3}}{a_{n-2}}
\end{array}
$$

Let $\frac{a_{n+1} + a_{n-1}}{a_{n}} = b_{n-1} \quad (n>1)$, then $b_{n} = b_{n-2}$. Given $b_{1} = b_{2} = 3$, so $b_{n} = 3$.
Thus, $\frac{a_{n+1} + a_{n-1}}{a_{n}} = 3$, which simplifies to $a_{n+1} = 3a_{n} - a_{n-1} \quad (n>1)$.
Using the characteristic equation and initial values $a_{1} = a_{2} = 1$,
we find that $a_{n} = \frac{5 - 2\sqrt{5}}{5} \left( \frac{3 + \sqrt{5}}{2} \right)^{n} + \frac{5 + 2\sqrt{5}}{5} \left( \frac{3 - \sqrt{5}}{2} \right)^{n}$.