$\begin{aligned} & \text {} \sin h x=\frac{-4}{3} \\ & \because \quad \cos ^2 h x=1+\sin ^2 h x=1+\frac{16}{9}=\frac{25}{9} \\ & \quad \cos h x=\frac{-5}{3} \\ & \sin h 2 x+\cos h 2 x \\ & =(2 \sin h x \cdot \cos h x)+\left(\cos ^2 h x+\sin h^2 x\right) \\ & =2\left(\frac{-4}{3}\right)\left(\frac{-5}{3}\right)+\left(\frac{25}{9}+\frac{16}{9}\right)=\frac{40}{9}+\frac{41}{9}=9 .\end{aligned}$