(i) We start with the identity \(\sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)\).

Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

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

Using the identity \(\sin^2 \theta - \cos^2 \theta = 2\sin^2 \theta - 1\), we get:

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

This completes the proof.

(ii) Given \(\sin^4 \theta - \cos^4 \theta = \frac{1}{2}\), we use the result from part (i):

\(2\sin^2 \theta - 1 = \frac{1}{2}\).

Solving for \(\sin^2 \theta\), we have:

\(2\sin^2 \theta = \frac{3}{2}\)

\(\sin^2 \theta = \frac{3}{4}\)

\(\sin \theta = \pm \frac{\sqrt{3}}{2}\).

Thus, \(\theta = 60^\circ, 120^\circ, 240^\circ, 300^\circ\).