(\frac{8}{5})^{-1}+(\frac{1}{2})^3*[\frac{2}{3}*5-(\frac{11}{29})^0]=
=\frac{5}{8}+\frac{1}{8}*[\frac{10}{3}-1]=\frac{5}{8}*\frac{1}{8}*[3\frac{1}{3}-1]==
=\frac{5}{8}-\frac{1}{8}*2\frac{1}{3}=\frac{5}{8}-\frac{1}{8}*\frac{7}{8}=
=\frac{5}{8}-\frac{7}{64}=\frac{40}{64}-\frac{7}{64}=\frac{33}{64}