(i) Use the angle subtraction and addition formulas:

\(\cos(3x-x) = \cos 2x = \cos 3x \cos x + \sin 3x \sin x\)

\(\cos(3x+x) = \cos 4x = \cos 3x \cos x - \sin 3x \sin x\)

Subtract these equations:

\(\cos 2x - \cos 4x = 2 \sin 3x \sin x\)

Thus, \(\frac{1}{2}(\cos 2x - \cos 4x) = \sin 3x \sin x\).

(ii) The integral becomes:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \sin 3x \sin x \, dx = \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \left( \frac{1}{4} \sin 2x - \frac{1}{8} \sin 4x \right) \, dx\)

Integrate each term:

\(\int \sin 2x \, dx = -\frac{1}{2} \cos 2x\)

\(\int \sin 4x \, dx = -\frac{1}{4} \cos 4x\)

Evaluate from \(\frac{1}{6}\pi\) to \(\frac{1}{3}\pi\):

\(\left[ -\frac{1}{8} \cos 2x + \frac{1}{32} \cos 4x \right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi}\)

Calculate the definite integral:

\(= \left( -\frac{1}{8} \cos \frac{2}{3}\pi + \frac{1}{32} \cos \frac{4}{3}\pi \right) - \left( -\frac{1}{8} \cos \frac{1}{3}\pi + \frac{1}{32} \cos \frac{2}{3}\pi \right)\)

\(= \frac{1}{8}\sqrt{3}\)