(i) Differentiate \(y = 8 \sin \frac{1}{2}x - \tan \frac{1}{2}x\) to find the maximum point. The derivative is:

\(4 \cos \frac{1}{2}x - \frac{1}{2} - \sec^2 \frac{1}{2}x\).

Set the derivative to zero:

\(4 \cos \frac{1}{2}x - \frac{1}{2} - \sec^2 \frac{1}{2}x = 0\).

Solve for \(\cos \frac{1}{2}x = \frac{1}{2}\).

This gives \(\alpha = \frac{2}{3}\pi\).

(ii) Integrate \(y = 8 \sin \frac{1}{2}x - \tan \frac{1}{2}x\) to find the area:

\(\int_0^{\frac{2}{3}\pi} \left( 8 \sin \frac{1}{2}x - \tan \frac{1}{2}x \right) \, dx\).

The integral is:

\(-16 \cos \frac{1}{2}x + 2 \ln \cos \frac{1}{2}x\).

Evaluate from \(0\) to \(\frac{2}{3}\pi\):

\(8 + 2 \ln \frac{1}{2}\).