$\text{a)}\sqrt{5^{-3}\cdot 5^5}=\sqrt{5^2}=5$   

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$\text{b)}\sqrt{6^7\cdot 6^{-3}}=\sqrt{6^4}=\sqrt{{\left(6^2\right)}^2}=6^2=36$  

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$\text{c)}\sqrt{{12}^8\cdot {\left(\frac{1}{6}\right)}^8}=\sqrt{{\left(12\cdot \frac{1}{6}\right)}^8}=\sqrt{2^8}=\sqrt{{\left(2^4\right)}^2}=2^4=16$  

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$\text{d)}\sqrt{{0,016}^2\cdot {0,1}^2}=\sqrt{{\left(0,016\cdot 0,1\right)}^2}=\sqrt{{\left(0,0016\right)}^2}=0,0016$