$a)$   $\frac{2^8-6^3}{2^7-5^3}=\frac{256-216}{128-125}=\frac{40}{3}$  

---

$b)$   $\left(2^{-3}+{\left(\frac{1}{2}\right)}^4\right)\cdot {\left(\frac{1}{4}\right)}^{-2}=\left(\frac{1}{2^3}+\frac{1}{16}\right)\cdot 4^2=\left(\frac{1}{8}+\frac{1}{16}\right)\cdot 16=\left(\frac{2}{16}+\frac{1}{16}\right)\cdot 16=$  

$=\frac{3}{16}\cdot 16=3$  

---

$c)$   ${\left(\frac{2}{3}\right)}^5\cdot {\left(\frac{2}{5}\right)}^{-4}\cdot {\left(\frac{5}{3}\right)}^{-3}=\frac{2^5}{3^5}\cdot {\left(\frac{5}{2}\right)}^4\cdot {\left(\frac{3}{5}\right)}^3=\frac{2^{\sout{5}}}{3^5}\cdot \frac{5^4}{\sout{2^4}}\cdot \frac{3^3}{5^3}=$  

$=\frac{2}{3^{{\sout{5}}^2}}\cdot \frac{5^4}{1}\cdot \frac{\sout{3^3}}{5^3}=\frac{2}{3^2}\cdot \frac{5^{\sout{4}}}{1}\cdot \frac{1}{\sout{5^3}}=\frac{2}{3^2}\cdot \frac{5}{1}\cdot \frac{1}{1}=\frac{2}{9}\cdot 5=\frac{10}{9}$