(i) We start with the given \(V = \frac{4}{3}h^3\). Differentiating with respect to \(t\), we have \(\frac{dV}{dt} = 4h^2 \frac{dh}{dt}\). The rate of liquid poured in is 20 m\(^3\) per hour, and the rate of leakage is \(kh^2\). Thus, \(\frac{dV}{dt} = 20 - kh^2\). Equating the two expressions for \(\frac{dV}{dt}\), we have:

\(4h^2 \frac{dh}{dt} = 20 - kh^2\)

When \(h = 1\), \(\frac{dh}{dt} = 4.95\), so:

\(4(1)^2(4.95) = 20 - k(1)^2\)

\(19.8 = 20 - k\)

\(k = 0.2\)

Substituting \(k = 0.2\) back, we get:

\(4h^2 \frac{dh}{dt} = 20 - 0.2h^2\)

\(\frac{dh}{dt} = \frac{5}{h^2} - \frac{1}{20}\)

(ii) We need to verify:

\(\frac{20h^2}{100-h^2} = -20 + \frac{2000}{(10-h)(10+h)}\)

Expanding the right side:

\(\frac{2000}{(10-h)(10+h)} = \frac{2000}{100-h^2}\)

Thus, \(-20 + \frac{2000}{100-h^2} = \frac{2000 - 20(100-h^2)}{100-h^2} = \frac{20h^2}{100-h^2}\)

(iii) Separating variables in the differential equation:

\(\frac{dh}{\frac{5}{h^2} - \frac{1}{20}} = dt\)

Integrating both sides, we have:

\(\int \frac{h^2}{5 - \frac{h^2}{20}} \, dh = \int dt\)

\(\int \frac{20h^2}{100-h^2} \, dh = t + C\)

Using partial fraction decomposition:

\(\frac{20h^2}{100-h^2} = -20 + \frac{2000}{(10-h)(10+h)}\)

Integrating gives:

\(-20h + 100 \ln(10+h) + 100 \ln(10-h) = t + C\)

\(t = 100 \ln \left( \frac{10+h}{10-h} \right) + C\)