$a)5^{-4}\cdot {16}^{-1}-{0,01}^2={\left(\frac{1}{5}\right)}^4\cdot \frac{1}{16}-{\left(\frac{1}{100}\right)}^2=\frac{1}{625}\cdot \frac{1}{16}-\frac{1}{10000}=\frac{1}{10000}-\frac{1}{10000}=0$   $b)\frac{{10}^{-3}+{10}^{-1}+10}{{\left(3^3\right)}^3:3^{2^3}}=$ $b)\frac{{10}^{-3}+{10}^{-1}+10}{{\left(3^3\right)}^3:3^{2^3}}=$   $\frac{0,001+0,1+10}{3^9:3^8}=\frac{10,101}{3^1}=\frac{10,101}{3}=3,367$  

$c)\frac{{0,1}^{-4}+{0,1}^{-2}+{0,1}^0}{\left(3^{-3}+3^{-3}+3^{-3}\right)\cdot \left({\left(\frac{1}{6}\right)}^{-2}-{\left(\frac{1}{3}\right)}^{-2}\right)}=$   $\frac{{\left(\frac{1}{10}\right)}^{-4}+{\left(\frac{1}{10}\right)}^{-2}+1}{\left(3\cdot 3^{-3}\right)-\left(6^2+3^2\right)}=$   $\frac{{10}^4+{10}^2+1}{\left(3^1\cdot 3^{-3}\right)\cdot \left(6^2-3^2\right)}=$  

$=\frac{10000+100+1}{3^{1+\left(-3\right)}\cdot \left(36-9\right)}=\frac{100101}{3^{-2}\cdot 27}=\frac{100101}{3^{-2}\cdot 3^3}=\frac{100101}{3^1}=\frac{100101}{3}=3367$