Algebra

Note that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2b^2} =\frac{(a+b)^2 - 2ab} {(ab)^2}$. By Vieta's, $a+b = \frac{-(-11)}{5} = \frac{11}{5}$ and $ab = \frac{4}{5}$, so $$\frac{(a+b)^2 - 2ab}{(ab)^2} = \frac{\frac{121}{25} - 2 \cdot \frac{4}{5}}{\left(\frac{4}{5}\right)^2} = \frac{\frac{81}{25}}{\frac{16}{25}} = \boxed{\frac{81}{16}}.$$