$\begin{aligned} & \text { } \operatorname{sech}^2\left(\tanh ^{-1} \frac{1}{2}\right)+\operatorname{cosech}^2\left(\cot h^{-1}(3)\right) \\ & =\operatorname{sech}^2\left(\operatorname{sech}^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)+\operatorname{cosech}^2\left(\operatorname{cosech}^{-1} \sqrt{8}\right) \\ & =\frac{3}{4}+8=\frac{35}{4}\end{aligned}$