(a) \[\sin4x=2\sin2x\cos2x=4\sin x\cos x(2\cos^2x-1)=4\sin x(2\cos^3x-\cos x).\]

(b) By the identity, the integrand is \(8\sin x\cos^6x-4\sin x\cos^4x\). Thus \[\int_0^{\pi/4}\cos^3x\sin4x\,dx=\left[-\frac87\cos^7x+\frac45\cos^5x\right]_0^{\pi/4}.\] Using \(\cos(\pi/4)=1/\sqrt2\), the upper value is \(\sqrt2/35\), and the lower value is \(-12/35\). Therefore the integral is \[\frac{12+\sqrt2}{35}.\]