(i) To find the intersection point \(A(p, q)\), set the equations equal: \(11 - x^2 = 5 - x\).

Simplify to get \(x^2 - x - 6 = 0\).

Factor the quadratic: \((x + 2)(x - 3) = 0\).

The solutions are \(x = -2\) and \(x = 3\). Since \(p\) is positive, \(p = 3\).

Substitute \(x = 3\) into \(y = 5 - x\) to find \(q\):

\(q = 5 - 3 = 2\).

(ii) To find \(f^{-1}(x)\), solve each piece of \(f(x)\) for \(x\).

For \(y = 11 - x^2\), solve for \(x\):

\(x^2 = 11 - y\) gives \(x = \sqrt{11 - y}\).

This is valid for \(2 \leq y \leq 11\) since \(f(x) = 11 - x^2\) for \(0 \leq x \leq 3\).

For \(y = 5 - x\), solve for \(x\):

\(x = 5 - y\).

This is valid for \(y < 2\) since \(f(x) = 5 - x\) for \(x > 3\).

Thus, \(f^{-1}(x) = \begin{cases} \sqrt{11 - x} & \text{for } 2 \leq x \leq 11, \\ 5 - x & \text{for } x < 2. \end{cases}\)