2. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$
Given $a=\frac{-1+\sqrt{5}}{2}$, we can obtain the corresponding quadratic equation $a^{2}+a-1=0$.
Thus, the original expression is
$$
\begin{array}{l}
=\frac{\left(a^{2}+a-1\right)\left(a^{3}-a\right)-2(a-1)}{a\left(a^{2}-1\right)} \\
=\frac{-2}{a^{2}+a}=-2 .
\end{array}
$$