(i) The rate of increase in height is proportional to \((9 - h)^{\frac{1}{3}}\), so we can write \(\frac{dh}{dt} = k(9 - h)^{\frac{1}{3}}\). Given \(\frac{dh}{dt} = 0.2\) when \(h = 1\), we have \(0.2 = k(9 - 1)^{\frac{1}{3}}\). Solving for \(k\), we find \(k = 0.1\). Thus, \(\frac{dh}{dt} = 0.1(9 - h)^{\frac{1}{3}}\).

(ii) Separate variables: \(\frac{dh}{(9 - h)^{\frac{1}{3}}} = 0.1 \, dt\). Integrate both sides: \(\int (9 - h)^{-\frac{1}{3}} \, dh = \int 0.1 \, dt\). This gives \(-\frac{3}{2}(9 - h)^{\frac{2}{3}} = 0.1t + C\). Use initial conditions \(t = 0, h = 1\) to find \(C\): \(-\frac{3}{2}(9 - 1)^{\frac{2}{3}} = C\). Solve for \(h\): \(h = 9 - \left(4 - \frac{1}{15}t\right)^{\frac{3}{2}}\).

(iii) The maximum height occurs when \(\frac{dh}{dt} = 0\), which implies \(9 - h = 0\), so \(h = 9\). The time taken is when \(h = 9\), which is \(t = 60\) years.

(iv) For half the maximum height, \(h = \frac{9}{2}\). Substitute into the expression for \(h\): \(\frac{9}{2} = 9 - \left(4 - \frac{1}{15}t\right)^{\frac{3}{2}}\). Solve for \(t\) to find \(t \approx 19.1\) years.