$a\text{)}{\left(\frac{a^{-2}}{1-a^2}+\frac{b^{-2}}{1-b^{-2}}\right)}^{-2}={\left(\frac{\frac{1}{a^2}}{1-\frac{1}{a^2}}+\frac{\frac{1}{b^2}}{1-\frac{1}{b^2}}\right)}^{-2}=$  

$={\left(\frac{\frac{1}{{\left(\sqrt{2}\right)}^2}}{1-\frac{1}{{\left(\sqrt{2}\right)}^2}}+\frac{\frac{1}{{\left(\sqrt{3}\right)}^2}}{1-\frac{1}{{\left(\sqrt{3}\right)}^2}}\right)}^{-2}={\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}+\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)}^{-2}=$  

$={\left(\frac{\frac{1}{2}}{\frac{1}{2}}+\frac{\frac{1}{3}}{\frac{2}{3}}\right)}^{-2}={\left(1+\frac{1}{3}\cdot \frac{3}{2}\right)}^{-2}={\left(1+\frac{1}{2}\right)}^{-2}={\left(1\frac{1}{2}\right)}^{-2}=$  

$={\left(\frac{3}{2}\right)}^{-2}={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$  

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$b\text{)}{\left(1+a^{-1}\right)}^{-1}+{\left(1-a^{-1}\right)}^{-1}={\left[1+{\left(-\frac{1}{3}\right)}^{-1}\right]}^{-2}+{\left[1-{\left(-\frac{1}{3}\right)}^{-1}\right]}^{-2}=$  

$={\left(1-3\right)}^{-2}+{\left(1+3\right)}^{-2}={\left(-2\right)}^{-2}+4^{-2}={\left(-\frac{1}{2}\right)}^2+{\left(\frac{1}{4}\right)}^2=$  

$=\frac{1}{4}+\frac{1}{16}=\frac{4}{16}+\frac{1}{16}=\frac{5}{16}$