\(a,\tan B=\dfrac{8}{15}=\dfrac{AC}{AB}=\dfrac{AC}{30}\\ \Leftrightarrow AC=\dfrac{30\cdot8}{15}=16\left(cm\right)\\ \Leftrightarrow BC=\sqrt{AC^2+AB^2}=34\left(cm\right)\)
\(b,\sin B=\dfrac{AC}{BC}=\dfrac{16}{34}=\dfrac{8}{17}\\ \cos B=\dfrac{AB}{BC}=\dfrac{30}{34}=\dfrac{15}{17}\\ \cot B=\dfrac{1}{\tan B}=\dfrac{15}{8}\)