Verified answer

[text]1)\frac{b_{12}}{b_{8}}=\frac{4}{1}\\\\b_{12}=4b_{8}\\\\b_{1}*q^{11}=4b_{1}*q^{7} |:b_{1}q^{7}\\\\q^{4}=4\\\\q=\sqrt[4]{4}=\sqrt{2}

2)(b_{2}-b_{4})^{2}-b_{5}=-\frac{35}{18}\\\\(b_{1}q-b_{1}q^{3})^{2}-b_{1}q^{4}=-\frac{35}{18}\\\\(b_{1}q)^{2}*(1-q^{2})^{2}-b_{1}q^{4}+\frac{35}{18}=0\\\\(b_{1}*\sqrt{2})^{2}*(1-(\sqrt{2})^{2} )^{2}-b_{1}*(\sqrt{2})^{4}+\frac{35}{18}=0\\\\2b_{1}^{2}-4b_{1}+\frac{35}{18}=0\\\\36b_{1}^{2}-72b_{1}+35=0\\\\D=(-72)^{2}-4*36*35=5184-5040=144=12^{2}\\\\b_{1}'=\frac{72-12}{72}=\frac{60}{72}=\frac{5}{6}\\\\b_{1}''=\frac{72+12}{72}=\frac{84}{72}=1\frac{1}{6}>1

b_{7}=b_{1}*q^{6}=\frac{5}{6}*(\sqrt{2})^{6}=\frac{5}{6}*8=\frac{20}{3}=6\frac{2}{3}\\\\Otvet:\boxed{b_{7}=6\frac{2}{3}}[\text]