$\begin{array}{l} {a^{m} \cdot a^{n}=a^{m+n}} \\ \dfrac{a^{m}}{a^{n}}=a^{m-n} \\ a^{n}\cdot b^{n}=({ab})^{n} \\ (a^{n})^{m}=a^{n\cdot m} \\ \\ \textbf{Gegeben:} \\ {a=\frac{1}{4} \qquad b=\frac{1}{5} \qquad m=2 \qquad n=4}\\ \\ \textbf{Rechnung:} \\ {\left(\frac{1}{4}\right)^{2} \cdot \left(\frac{1}{4}\right)^{4}=\left(\frac{1}{4}\right)^{2+4}=\left(\frac{1}{4}\right)^{6}=0,000244}\\ \left(\frac{1}{4}\right)^{2}:\left(\frac{1}{4}\right)^{4}=\dfrac{\left(\frac{1}{4}\right)^{2}}{\left(\frac{1}{4}\right)^{4}}=\left(\frac{1}{4}\right)^{2-4}=\left(\frac{1}{4}\right)^{-2}=16\\ \left(\frac{1}{4}\right)^{4}\cdot \left(\frac{1}{5}\right)^{4}=(\frac{1}{4}\cdot\frac{1}{5})^{4}= \left(\frac{1}{20}\right)^{4}={6,25\cdot 10^{-6}} \\ (\left(\frac{1}{4}\right)^{4})^{2}=\left(\frac{1}{4}\right)^{4\cdot 2} = \left(\frac{1}{4}\right)^{8}={1,53\cdot 10^{-5}} \end{array}$