\(1,BC=\sqrt{AB^2+AC^2}=25\left(cm\right)\\ BH=\dfrac{AB^2}{BC}=9\left(cm\right)\\ \sin B=\dfrac{AC}{BC}=\dfrac{4}{5}\approx\sin53^0\Rightarrow\widehat{B}\approx53^0\\ 2,\cos^2x=1-\sin^2x=1-0,36=0,64\\ \Rightarrow\cos x=0,8\\ \Rightarrow\tan x=\dfrac{\sin x}{\cos x}=\dfrac{3}{4}\)