\[q = - \frac{1}{3}\]
\[\frac{b_{8} \cdot b_{10}}{b_{9} \cdot b_{11}} = \frac{b_{1}q^{7} \cdot b_{1}q^{9}}{b_{1}q^{8} \cdot b_{1}q^{10}} = \frac{q^{16}}{q^{18}} =\]
\[= \frac{1}{q^{2}} = \frac{1}{\left( - \frac{1}{3} \right)^{2}} = \frac{1}{\frac{1}{9}} = 9.\]
\[Ответ:9.\]