(i) For \(0 \leq x \leq 2\), \(f(x) = \frac{1}{2}x^2\). The minimum value is \(0\) and the maximum value is \(2\). Thus, the range is \(0 \leq f(x) \leq 2\).

For \(2 < x \leq 6\), \(f(x) = \frac{1}{2}x + 1\). The minimum value is \(2\) and the maximum value is \(4\). Thus, the range is \(2 < f(x) \leq 4\).

Combining both parts, the range of \(f\) is \(0 < f(x) < 4\).

(ii) To sketch \(y = f^{-1}(x)\), reflect the graph of \(f\) across the line \(y = x\).

(iii) For \(0 < x < 2\), solve \(y = \frac{1}{2}x^2\) for \(x\):

\(x = \sqrt{2y}\).

For \(2 < x < 4\), solve \(y = \frac{1}{2}x + 1\) for \(x\):

\(x = 2y - 2\).

Thus, \(f^{-1}(x)\) is defined by \(x \mapsto \sqrt{2x}\) for \(0 < x < 2\) and \(x \mapsto 2x - 2\) for \(2 < x < 4\).