\((1)\)已知\(\cos \beta =-\dfrac{3}{5}\)，且\(\beta \)为第二象限角，求\(\sin \beta \)的值；

\((2)\)已知\(\dfrac{\tan \alpha }{\tan \alpha -6}=-1,\)求值：\(\dfrac{2\cos \alpha -3\sin \alpha }{3\cos \alpha +4\sin \alpha }.\)