(i) To prove the identity \(\cos 4\theta - 4 \cos 2\theta + 3 \equiv 8 \sin^4 \theta\), we start by using the double angle formulas:

\(\cos 2\theta = 1 - 2\sin^2 \theta\)

\(\cos 4\theta = 2\cos^2 2\theta - 1 = 2(1 - 2\sin^2 \theta)^2 - 1\)

Expanding \(\cos 4\theta\):

\(\cos 4\theta = 2(1 - 4\sin^2 \theta + 4\sin^4 \theta) - 1\)

\(= 2 - 8\sin^2 \theta + 8\sin^4 \theta - 1\)

\(= 1 - 8\sin^2 \theta + 8\sin^4 \theta\)

Now substitute into the original expression:

\(\cos 4\theta - 4\cos 2\theta + 3 = (1 - 8\sin^2 \theta + 8\sin^4 \theta) - 4(1 - 2\sin^2 \theta) + 3\)

\(= 1 - 8\sin^2 \theta + 8\sin^4 \theta - 4 + 8\sin^2 \theta + 3\)

\(= 8\sin^4 \theta\)

Thus, the identity is proven.

(ii) Using the result from part (i), we find the integral:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \sin^4 \theta \, d\theta\)

Using the identity \(\sin^4 \theta = \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)\), the integral becomes:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3) \, d\theta\)

\(= \frac{1}{8} \left[ \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \cos 4\theta \, d\theta - 4 \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \cos 2\theta \, d\theta + 3 \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} 1 \, d\theta \right]\)

Calculate each integral:

\(\int \cos 4\theta \, d\theta = \frac{1}{4} \sin 4\theta\)

\(\int \cos 2\theta \, d\theta = \frac{1}{2} \sin 2\theta\)

\(\int 1 \, d\theta = \theta\)

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

\(\frac{1}{8} \left[ \frac{1}{4} (\sin \frac{4}{3}\pi - \sin \frac{2}{3}\pi) - 4 \times \frac{1}{2} (\sin \frac{2}{3}\pi - \sin \frac{1}{3}\pi) + 3 \left( \frac{1}{3}\pi - \frac{1}{6}\pi \right) \right]\)

\(= \frac{1}{8} \left[ 0 - 2(\sqrt{3}/2 - 1/2) + \frac{\pi}{2} \right]\)

\(= \frac{1}{8} \left[ -\sqrt{3} + \frac{\pi}{2} \right]\)

\(= \frac{1}{32} (2\pi - \sqrt{3})\)