(i) The length of AS is \(r \tan \theta\). The area of triangle OAB is \(r^2 \tan \theta\) or \(\frac{1}{2} r^2 \tan \theta\). The area of sector OPS is \(\frac{1}{2} r^2 \times 2\theta = r^2 \theta\). Therefore, the shaded area is \(r^2 (\tan \theta - \theta)\).

(ii) Given \(\theta = \frac{1}{3}\pi\) and \(r = 6\), we have \(\cos \frac{\pi}{3} = \frac{6}{OA} \Rightarrow OA = 12\). Thus, \(AP = 6\). The length of AS is \(6 \tan \frac{\pi}{3} \Rightarrow AB = 12\sqrt{3}\). The arc PST is \(12 \times \frac{\pi}{3}\). Therefore, the perimeter is \(12 + 12\sqrt{3} + 4\pi\).