$\begin{array}{l} {a^{m} \cdot a^{n}=a^{m+n}} \\ \dfrac{a^{m}}{a^{n}}=a^{m-n} \\ a^{n}\cdot b^{n}=({ab})^{n} \\ (a^{n})^{m}=a^{n\cdot m} \\ \\ \textbf{Gegeben:} \\ {a=\frac{1}{2} \qquad b=\frac{1}{5} \qquad m=3 \qquad n=4}\\ \\ \textbf{Rechnung:} \\ {\left(\frac{1}{2}\right)^{3} \cdot \left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{3+4}=\left(\frac{1}{2}\right)^{7}=0,00781}\\ \left(\frac{1}{2}\right)^{3}:\left(\frac{1}{2}\right)^{4}=\dfrac{\left(\frac{1}{2}\right)^{3}}{\left(\frac{1}{2}\right)^{4}}=\left(\frac{1}{2}\right)^{3-4}=\left(\frac{1}{2}\right)^{-1}=2\\ \left(\frac{1}{2}\right)^{4}\cdot \left(\frac{1}{5}\right)^{4}=(\frac{1}{2}\cdot\frac{1}{5})^{4}= \left(\frac{1}{10}\right)^{4}={0,0001} \\ (\left(\frac{1}{2}\right)^{4})^{3}=\left(\frac{1}{2}\right)^{4\cdot 3} = \left(\frac{1}{2}\right)^{12}={0,000244} \end{array}$