(i) Start with the left-hand side: \(\frac{2 \sin x - \sin 2x}{1 - \cos 2x}\).

Using the double angle formula, \(\sin 2x = 2 \sin x \cos x\) and \(\cos 2x = 2 \cos^2 x - 1\), substitute these into the expression:

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

Simplify the denominator: \(2 - 2 \cos^2 x = 2(1 - \cos^2 x) = 2 \sin^2 x\).

Thus, the expression becomes \(\frac{2 \sin x (1 - \cos x)}{2 \sin^2 x} = \frac{\sin x (1 - \cos x)}{\sin^2 x} = \frac{\sin x}{1 + \cos x}\), which matches the right-hand side.

(ii) The integral \(\int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \frac{2 \sin x - \sin 2x}{1 - \cos 2x} \, dx\) can be rewritten using the result from part (i) as \(\int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \frac{\sin x}{1 + \cos x} \, dx\).

Let \(u = 1 + \cos x\), then \(du = -\sin x \, dx\), so \(dx = -\frac{du}{\sin x}\).

The integral becomes \(-\int \frac{1}{u} \, du = -\ln |u| + C\).

Substitute back \(u = 1 + \cos x\): \(-\ln |1 + \cos x|\).

Evaluate from \(x = \frac{1}{3}\pi\) to \(x = \frac{1}{2}\pi\):

\(-\ln |1 + \cos \frac{1}{2}\pi| + \ln |1 + \cos \frac{1}{3}\pi|\).

\(\cos \frac{1}{2}\pi = 0\) and \(\cos \frac{1}{3}\pi = \frac{1}{2}\), so the expression becomes:

\(-\ln(1) + \ln\left(1 + \frac{1}{2}\right) = \ln\left(\frac{3}{2}\right)\).