a,
$\dfrac{\pi}{2}<a<\pi\to \sin a>0,\cos a<0$
$\sin a=\dfrac{3}{5}\to \cos a=-\sqrt{1-\sin^2a}=\dfrac{-4}{5}$
$\cos\Big(a+\dfrac{\pi}{3}\Big)=\cos a\cos\dfrac{\pi}{3}-\sin a\sin\dfrac{\pi}{3}=\dfrac{-4-3\sqrt3}{10}$
b,
$\sin2b=2\sin b\cos b=0,6$
$\sin b=0,3\to \cos b=1$
Vì $0,3^2+1^2\ne 1$ nên hai đẳng thức $\sin2b=0,6$ và $\sin b=0,3$ không xảy ra cùng nhau.
c,
$VT=\sin x\cos 4x-\sin2x\cos3x$
$=\dfrac{1}{2}\Big(\sin5x+\sin(-3x)\Big)-\dfrac{1}{2}\Big(\sin5x+\sin(-x)\Big)$
$=\dfrac{1}{2}\sin5x-\dfrac{1}{2}\sin3x-\dfrac{1}{2}\sin5x+\dfrac{1}{2}\sin x$
$=\dfrac{1}{2}\Big(\sin x-\sin3x\Big)$
$=\cos2x.\sin(-x)$
$=-\sin x(1-2\sin^2x)$
$=2\sin^3x-\sin x$
$=VP$