\[\boxed{\text{633.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ b_{1} = 125,\ \ b_{3} = b_{1} \cdot q^{2} = 5,\ \ 125q² = 5,\]
\[q² = \frac{1}{25},\ \ q = \pm \frac{1}{5},\ \ b_{6} = 125 \cdot \left( \pm \frac{1}{5} \right)^{5} = \pm \frac{5^{3}}{5^{5}} = \pm \frac{1}{25};\ \]
\[\textbf{б)}\ b_{1} = - \frac{2}{9},\ \ b_{3} = b_{1} \cdot q^{2} = - 2,\ \ - \frac{2}{9q^{2}} = - 2,\]
\[q² = 9,\ \ q = \pm 3;\]
\[b_{7} = b_{1} \cdot q^{6} = - \frac{2}{9} \cdot ( \pm 3)^{6} = - \frac{2}{3^{2}} \cdot 3^{6} = - 2 \cdot 3^{4} = - 162;\]
\[\textbf{в)}\ b_{4} = b_{1} \cdot q³ = - 1,\ \ b_{6} = b_{1} \cdot q^{5} = - 100,\ \ \frac{b_{1}q^{5}}{b_{1}q^{3}} = \frac{- 100}{- 1} = 100;\]
\[q^{2} = 100,\ \ q = \pm 10;\]
\[b_{1} = \frac{b_{4}}{q^{3}} = \frac{- 1}{\pm 10³} = \pm 0,001.\]
\[Ответ:\ а)\ \frac{1}{25}\ \ или\ \left( - \frac{1}{25} \right);\ \ б) - 162;\ \ в) - 0,001\ или\ 0,001.\]