(i) For \(\theta = 1\), angle BOC = \(\pi - \theta\). The area of sector BOC is given by \(\frac{1}{2} r^2 (\pi - \theta)\). Substituting \(r = 8\) and \(\theta = 1\), the area is \(\frac{1}{2} \times 8^2 \times (\pi - 1) = 32(\pi - 1)\) or approximately 68.5.

(ii) The perimeter of sector AOB is \(8 + 8 + 8\theta\) and the perimeter of sector BOC is \(8 + 8 + 8(\pi - \theta)\). Setting \(8 + 8 + 8\theta = \frac{1}{2}(8 + 8 + 8(\pi - \theta))\), solve for \(\theta\):

\(16 + 8\theta = 8 + 8 + 4\pi - 4\theta\)

\(12\theta = 4\pi - 16\)

\(\theta = \frac{1}{3}(\pi - 2)\) or approximately 0.381.

(iii) For \(\theta = \frac{1}{3}\pi\), AB = 8 cm. Using the cosine rule or trigonometry, BC = \(2 \times 8 \sin \frac{\pi}{3} = 8\sqrt{3}\). The perimeter of triangle ABC is \(8 + 8 + 8\sqrt{3} = 24 + 8\sqrt{3}\) cm.