a)

$x=\frac{{27}^9+{81}^7}{9^{13}}=\frac{{\left(3^3\right)}^9+{\left(3^4\right)}^7}{{\left(3^2\right)}^{13}}=\frac{3^{27}+3^{28}}{3^{26}}=$     

$=\frac{3^{27}}{3^{26}}+\frac{3^{28}}{3^{26}}=3^1+3^2=3+9=12$  

 

$y=\frac{8^{16}+4^{25}}{5\cdot {16}^{12}}=\frac{{\left(2^3\right)}^{16}+{\left(2^2\right)}^{25}}{5\cdot {\left(2^4\right)}^{12}}=\frac{2^{48}+2^{50}}{5\cdot 2^{48}}=$  

$=\frac{2^{48}}{5\cdot 2^{48}}+\frac{2^{50}}{5\cdot 2^{48}}=\frac{1}{5}+\frac{2^2}{5}=\frac{1}{5}+\frac{4}{5}=1$  

 

Obliczmy jakim procentem liczby x jest liczba y.

$\frac{y}{x}\cdot 100\%=\frac{1}{12}\cdot 100\%=\frac{100}{12}\%=8\frac{4}{12}\%=8\frac{1}{3}\%$  

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b)

$x=\frac{5^{-20}+5^{-19}}{{125}^{-6}}=\frac{5^{-20}+5^{-19}}{{\left(5^3\right)}^{-6}}=\frac{5^{-20}+5^{-19}}{5^{-18}}=$  

$=\frac{5^{-20}}{5^{-18}}+\frac{5^{-19}}{5^{-18}}=5^{-2}+5^{-1}=\frac{1}{25}+\frac{1}{5}=\frac{1}{25}+\frac{5}{25}=\frac{6}{25}$  

 

$y=\frac{6^{-14}+12\cdot {\left(\frac{1}{6}\right)}^{15}}{{0,5}^{14}\cdot {\left(\frac{1}{3}\right)}^{15}}=\frac{6^{-14}+12\cdot 6^{-15}}{{\left(\frac{1}{2}\right)}^{14}\cdot {\left(\frac{1}{3}\right)}^{14}\cdot \frac{1}{3}}=\frac{6^{-14}\cdot \left(1+12\cdot 6^{-1}\right)}{{\left(\frac{1}{2}\cdot \frac{1}{3}\right)}^{14}\cdot \frac{1}{3}}=$  

$=\frac{6^{-14}\cdot \left(1+12\cdot \frac{1}{6}\right)}{{\left(\frac{1}{6}\right)}^{14}\cdot \frac{1}{3}}=\frac{6^{-14}\cdot \left(1+2\right)}{6^{-14}\cdot \frac{1}{3}}=\frac{3}{\frac{1}{3}}=3\cdot 3=9$  

 

Obliczmy jakim procentem liczby x jest liczba y.

$\frac{y}{x}\cdot 100\%=\frac{9}{\frac{6}{25}}\cdot 100\%={\cancel{9}}^3\cdot \frac{25}{{\cancel{6}}^2}\cdot 100\%=3\cdot \frac{25}{2}\cdot 100\%=$  

$=\frac{75}{2}\cdot 100\%=75\cdot 50\%=3750\%$