a)
[(1\frac{1}{3})^{-1}-2^{-2}]^{-3}=[(\frac{4}{3})^{-1}-(\frac{1}{2})^2]^{-3}=(\frac{3}{4}-\frac{1}{4})^{-3}=(\frac{2}{4})^{-3}=(\frac{4}{2})^3=2^3=8
b)
\frac{3*(\frac{2}{3})^{-2}+4^{-1}}{(0,2)^{-1}-(\frac{1}{2})^{-2}}=\frac{3*(\frac{3}{2})^2+\frac{1}{4}}{(\frac{2}{10})^{-1}-2^2}=\frac{3*\frac{9}{4}+\frac{1}{4}}{\frac{10}{2}-4}=\frac{\frac{27}{4}+\frac{1}{4}}{5-4}=\frac{\frac{28}{4}}{1}=7