$\begin{array}{l} {a^{m} \cdot a^{n}=a^{m+n}} \\ \dfrac{a^{m}}{a^{n}}=a^{m-n} \\ a^{n}\cdot b^{n}=({ab})^{n} \\ (a^{n})^{m}=a^{n\cdot m} \\ \\ \textbf{Gegeben:} \\ {a=-\frac{1}{3} \qquad b=2 \qquad m=2 \qquad n=2}\\ \\ \textbf{Rechnung:} \\ {\left(-\frac{1}{3}\right)^{2} \cdot \left(-\frac{1}{3}\right)^{2}=\left(-\frac{1}{3}\right)^{2+2}=\left(-\frac{1}{3}\right)^{4}=\frac{1}{81}}\\ \left(-\frac{1}{3}\right)^{2}:\left(-\frac{1}{3}\right)^{2}=\dfrac{\left(-\frac{1}{3}\right)^{2}}{\left(-\frac{1}{3}\right)^{2}}=\left(-\frac{1}{3}\right)^{2-2}=\left(-\frac{1}{3}\right)^{0}=1\\ \left(-\frac{1}{3}\right)^{2}\cdot 2^{2}=(\left(-\frac{1}{3}\right)\cdot2)^{2}= \left(-\frac{2}{3}\right)^{2}={\frac{4}{9}} \\ (\left(-\frac{1}{3}\right)^{2})^{2}=\left(-\frac{1}{3}\right)^{2\cdot 2} = \left(-\frac{1}{3}\right)^{4}={\frac{1}{81}} \end{array}$