\(a,BD=\sqrt{AB^2+AD^2}=5\left(cm\right)\\ \sin B=\dfrac{AD}{BD}=\dfrac{4}{5}\approx\sin53^0\Rightarrow\widehat{B}\approx53^0\\ \Rightarrow\widehat{D}\approx90^0-53^0=37^0\\ b,\left\{{}\begin{matrix}BM\cdot BD=AB^2\\AM\cdot AC=AB^2\end{matrix}\right.\left(HTL\right)\\ \Rightarrow BM\cdot BD=AM\cdot AC\)