To calculate generating function If $a_{n}$ = $\frac {1}{(n-1)(n+1)}$ for $n\ge2$
What are we supposed to do with $a_{0}$ and $a_{1}$?
How can I find the generating function without using $a_{0}$ and $a_{1}$?

Since $a_0$ and $a_1$ are not specified, one usually leaves them out of the generating function, essentially assuming they are $0$.
$$
\begin{align}
\sum_{k=2}^\infty\frac{x^k}{(k-1)(k+1)}
&=\frac12\sum_{k=2}^\infty\left(\frac1{k-1}-\frac1{k+1}\right)x^k\\
&=\frac12\sum_{k=1}^\infty\frac{x^{k+1}}{k}-\frac12\sum_{k=3}^\infty\frac{x^{k-1}}{k}\\
&=\frac12\left(x-\frac1x\right)\sum_{k=1}^\infty\frac{x^k}{k}+\frac12\left(1+\frac x2\right)\\
&=\frac{1-x^2}{2x}\log(1-x)+\frac{2+x}4
\end{align}
$$
