(i) The rate of change of volume \(V\) with respect to time \(t\) is given by:

\(\frac{dV}{dt} = 1 - 0.2\sqrt{h}.\)

The volume \(V\) is related to the depth \(h\) by \(V = 2h\), so:

\(\frac{dV}{dt} = 2 \frac{dh}{dt}.\)

Equating the two expressions for \(\frac{dV}{dt}\), we have:

\(2 \frac{dh}{dt} = 1 - 0.2\sqrt{h}.\)

Rearranging gives:

\(\frac{dh}{dt} = \frac{1 - 0.2\sqrt{h}}{2}.\)

To find \(T\), integrate with respect to \(h\) from 0 to 4:

\(T = \int_0^4 \frac{10}{5 - \sqrt{h}} \, dh.\)

(ii) Using the substitution \(u = 5 - \sqrt{h}\), we have:

\(\frac{du}{dh} = -\frac{1}{2\sqrt{h}} \Rightarrow du = -\frac{1}{2\sqrt{h}} \, dh.\)

Substitute for \(h\) and \(dh\):

\(T = \int_3^5 \frac{20(5-u)}{u} \, du.\)

Integrate to obtain:

\(T = 100\ln u - 20u \bigg|_3^5.\)

Substitute the limits:

\(T = (100\ln 5 - 100\ln 3) - (20 \times 5 - 20 \times 3).\)

Calculate to find \(T = 11.1\).