\(a,AC=\tan B\cdot AB=\dfrac{8}{15}\cdot30=16\left(cm\right)\\ BC=\sqrt{AB^2+AC^2}=34\left(cm\right)\left(pytago\right)\\ b,\sin B=\dfrac{AC}{BC}=\dfrac{16}{34}=\dfrac{8}{17}\\ \cos B=\dfrac{AB}{BC}=\dfrac{30}{34}=\dfrac{15}{17}\\ \cot B=\dfrac{1}{\tan B}=\dfrac{15}{8}\)