$\text{a)}{4}^{-3}\cdot {0,125}^{-3}={\left(4\cdot 0,125\right)}^{-3}={0,5}^{-3}={\left(\frac{1}{2}\right)}^{-3}=2^3=8$  

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$\text{b)}{2,5}^{-2}\cdot {0,2}^{-2}\cdot {\left(\frac{2}{3}\right)}^{-2}={\left(2,5\cdot 0,2\cdot \frac{2}{3}\right)}^{-2}={\left(0,5\cdot \frac{2}{3}\right)}^{-2}={\left(\frac{1}{2}\cdot \frac{2}{3}\right)}^{-2}={\left(\frac{1}{3}\right)}^{-2}=3^2=9$  

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$\text{c)}{\left(\frac{2}{3}\right)}^{-4}\cdot {\left(\frac{3}{4}\right)}^{-4}={\left(\frac{2}{3}\cdot \frac{3}{4}\right)}^{-4}={\left(\frac{1}{2}\right)}^{-4}=2^4=16$  

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$\text{d)}{2}^{-3}\cdot 3^{-3}\cdot {\left(\frac{1}{12}\right)}^{-3}={\left(2\cdot 3\cdot \frac{1}{12}\right)}^{-3}={\left(\frac{1}{2}\right)}^{-3}=2^3=8$  

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$\text{e)}{5}^{-7}\cdot 4^{-7}\cdot {0,05}^{-7}={\left(5\cdot 4\cdot 0,05\right)}^{-7}={\left(20\cdot 0,05\right)}^{-7}=1^{-7}=1$  

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$\text{f)}{2}^0\cdot 5^0:{\left(\frac{1}{3}\right)}^0=1\cdot 1:1=1$