(i) The rate of increase of the surface area is proportional to the volume, so \(\frac{dA}{dt} = kV\). Given \(A = 4\pi r^2\) and \(V = \frac{4}{3}\pi r^3\), differentiate \(A\) with respect to \(t\):

\(\frac{dA}{dt} = 8\pi r \frac{dr}{dt}\).

Equating \(8\pi r \frac{dr}{dt} = k \frac{4}{3}\pi r^3\), solve for \(\frac{dr}{dt}\):

\(\frac{dr}{dt} = \frac{k}{6} r^2\).

Using \(\frac{dr}{dt} = 2\) and \(r = 5\), find \(k\):

\(2 = \frac{k}{6} \times 25 \Rightarrow k = 0.48\).

Thus, \(\frac{dr}{dt} = 0.08r^2\).

(ii) Separate variables and integrate:

\(\int \frac{1}{r^2} \, dr = \int 0.08 \, dt\).

\(-\frac{1}{r} = 0.08t + C\).

Using \(t = 0, r = 5\), find \(C\):

\(-\frac{1}{5} = C\).

Thus, \(-\frac{1}{r} = 0.08t - \frac{1}{5}\).

Solve for \(r\):

\(r = \frac{5}{1 - 0.4t}\).

(iii) For \(r\) to be positive, \(1 - 0.4t > 0\), so \(t < 2.5\). Therefore, \(0 \leq t < 2.5\).