(i) The perimeter of the sector is given by the sum of the two radii and the arc length: \(2r + r\theta = 20\). Solving for \(\theta\), we have:

\(20 = 2r + r\theta\)

\(r\theta = 20 - 2r\)

\(\theta = \frac{20}{r} - 2\)

(ii) The area of the sector is given by \(\frac{1}{2}r^2\theta\). Substituting \(\theta = \frac{20}{r} - 2\), we get:

\(A = \frac{1}{2}r^2\left(\frac{20}{r} - 2\right)\)

\(A = \frac{1}{2}(20r - 2r^2)\)

\(A = 10r - r^2\)

(iii) Using the cosine rule in triangle \(OPQ\), where \(r = 8\) and \(\theta = 0.5\) radians:

\(PQ^2 = 8^2 + 8^2 - 2 \times 8 \times 8 \times \cos(0.5)\)

\(PQ^2 = 64 + 64 - 128 \times \cos(0.5)\)

\(PQ = \sqrt{128 - 128 \times \cos(0.5)}\)

Alternatively, using the sine rule:

\(PQ = 2 \times 8 \times \sin(0.25)\)

\(PQ \approx 3.96\) (allow 3.95)