1. Given $\operatorname{ctg} \theta=\sqrt[3]{7}\left(0^{\circ}<\theta<90^{\circ}\right)$. Then, $\frac{\sin ^{2} \theta+\sin \theta \cos \theta+2 \cos ^{2} \theta}{\sin ^{2} \theta+\sin \theta \cos \theta+\cos ^{2} \theta}=$ $\qquad$ .
\begin{aligned} \text { 2.1. } & \frac{13-\sqrt[3]{49}}{6} . \\ \text { Original expression } & =1+\frac{\cos ^{2} \theta}{\sin ^{2} \theta+\sin \theta \cos \theta+\cos ^{2} \theta} \\ & =1+\frac{\operatorname{ctg}^{2} \theta}{1+\operatorname{ctg} \theta+\operatorname{ctg}^{2} \theta} \\ & =1+\frac{\sqrt[3]{49}(\sqrt[3]{7}-1)}{(1+\sqrt[3]{7}+\sqrt[3]{49})(\sqrt[3]{7}-1)} \\ & =1+\frac{7-\sqrt[3]{49}}{6}=\frac{13-\sqrt[3]{49}}{6} .\end{aligned}