$a)2^{-3}\cdot 5^{-3}={\left(2\cdot 5\right)}^{-3}={10}^{-3}=\frac{1}{{10}^3}=\frac{1}{1000}$

$b)4^{-2}\cdot {15}^{-2}={\left(4\cdot 15\right)}^{-2}={60}^{-2}=\frac{1}{{60}^2}=\frac{1}{3600}$ $b)4^{-2}\cdot {15}^{-2}={\left(4\cdot 15\right)}^{-2}={60}^{-2}=\frac{1}{{60}^2}=\frac{1}{3600}$

$c){\left(\frac{1}{4}\right)}^{-5}\cdot 8^{-5}={\left(\frac{1}{4}\cdot 8\right)}^{-5}=2^{-5}=\frac{1}{2^5}=\frac{1}{32}$

$d){2,8}^{-1}\cdot {10}^{-1}={\left(2,8\cdot 10\right)}^{-1}={28}^{-1}=\frac{1}{28}$

$e){\left(\frac{3}{5}\right)}^{-2}\cdot {\left(1\frac{2}{3}\right)}^{-2}={\left(\frac{3}{5}\cdot 1\frac{2}{3}\right)}^{-2}={\left(\frac{3}{5}\cdot \frac{5}{3}\right)}^{-2}=1^{-2}=\frac{1}{1^2}=1$

$f){\left(-0,15\right)}^{-3}\cdot 6^{-3}={\left(-0,15\cdot 6\right)}^{-3}={\left(-0,9\right)}^{-3}={\left(-\frac{9}{10}\right)}^{-3}={\left(-\frac{10}{9}\right)}^3=-\frac{1000}{729}$