1. Given that $a$, $b$, and $c$ are non-zero rational numbers, and satisfy
$$
\begin{array}{l}
a b^{2}=\frac{c}{a}-b . \\
\text { Then }\left(\frac{a^{2} b^{2}}{c^{2}}-\frac{2}{c}+\frac{1}{a^{2} b^{2}}+\frac{2 a b}{c^{2}}-\frac{2}{a b c}\right) \div \\
\left(\frac{2}{a b}-\frac{2 a b}{c}\right) \div \frac{101}{c}=
\end{array}
$$
2. 1. $-\frac{1}{202}$.
From $a b^{2}=\frac{c}{a}-b$, we get
$$
\frac{c}{a b}-a b=1 \text {. }
$$

Notice that
$$
\begin{array}{l}
\frac{a^{2} b^{2}}{c^{2}}-\frac{2}{c}+\frac{1}{a^{2} b^{2}}=\left(\frac{a b}{c}-\frac{1}{a b}\right)^{2} . \\
\text { Then the original expression }=\left[\frac{1}{2}\left(\frac{1}{a b}-\frac{a b}{c}\right)-\frac{1}{c}\right] \div \frac{101}{c} \\
=\frac{1}{202}\left(\frac{c}{a b}-a b-2\right)=-\frac{1}{202} .
\end{array}
$$