$\begin{array}{l} {a^{m} \cdot a^{n}=a^{m+n}} \\ \dfrac{a^{m}}{a^{n}}=a^{m-n} \\ a^{n}\cdot b^{n}=({ab})^{n} \\ (a^{n})^{m}=a^{n\cdot m} \\ \\ \textbf{Gegeben:} \\ {a=\frac{1}{2} \qquad b=\frac{1}{6} \qquad m=2 \qquad n=3}\\ \\ \textbf{Rechnung:} \\ {\left(\frac{1}{2}\right)^{2} \cdot \left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{2+3}=\left(\frac{1}{2}\right)^{5}=\frac{1}{32}}\\ \left(\frac{1}{2}\right)^{2}:\left(\frac{1}{2}\right)^{3}=\dfrac{\left(\frac{1}{2}\right)^{2}}{\left(\frac{1}{2}\right)^{3}}=\left(\frac{1}{2}\right)^{2-3}=\left(\frac{1}{2}\right)^{-1}=2\\ \left(\frac{1}{2}\right)^{3}\cdot \left(\frac{1}{6}\right)^{3}=(\frac{1}{2}\cdot\frac{1}{6})^{3}= \left(\frac{1}{12}\right)^{3}={0,000579} \\ (\left(\frac{1}{2}\right)^{3})^{2}=\left(\frac{1}{2}\right)^{3\cdot 2} = \left(\frac{1}{2}\right)^{6}={\frac{1}{64}} \end{array}$