+26 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 \[\frac{(a+b)^2 - 2ab}{(ab)^2} = \frac{\frac{121}{25} - 2 \cdot \frac{4}{5}}{\left(\frac{4}{5}\right)^2} = \frac{\frac{81}{25}}{\frac{16}{25}} = \boxed{\frac{81}{16}}.\]
