\[b_{4} = b_{1} \cdot q^{3} = 9;\ \ \]
\[b_{6} = b_{1} \cdot q^{5} = \frac{1}{81}\]
\[\frac{b_{6}}{b_{4}} = q^{2} = \frac{1}{81 \cdot 9} = \frac{1}{729}\]
\[q = \pm \frac{1}{27}.\]
\[q = \frac{1}{27}:\ \]
\[b_{5} = b_{4} \cdot q = 9 \cdot \frac{1}{27} = \frac{1}{3}.\]
\[Ответ:q = \frac{1}{27};\ \ b_{5} = \frac{1}{3}.\]