${\left[{\left(\frac{5}{7}\right)}^{-3}\right]}^0+3\cdot 4^{-\frac{3}{2}}+{64}^{\frac{4}{3}}-{\left(-2\right)}^3=1+3\cdot {\left(2^2\right)}^{-\frac{3}{2}}+{\left(4^3\right)}^{\frac{4}{3}}-{\left(-2\right)}^3=$      

$=1+3\cdot 2^{2\cdot \frac{-3}{2}}+4^{3\cdot \frac{4}{3}}-\left(-8\right)=1+3\cdot 2^{-3}+4^4+8=$  

$=1+3\cdot {\left(\frac{1}{2}\right)}^3+256+8=1+\frac{3}{8}+256+8=265\frac{3}{8}$