(i) The rate of increase of m is proportional to \((50 - m)^2\), so \(\frac{dm}{dt} = k(50 - m)^2\). Given \(\frac{dm}{dt} = 5\) when \(m = 0\), we have \(5 = k(50)^2\). Solving for \(k\), we find \(k = \frac{5}{2500} = 0.002\). Thus, \(\frac{dm}{dt} = 0.002(50 - m)^2\).

(ii) Separate variables: \(\frac{1}{(50 - m)^2} dm = 0.002 dt\). Integrate both sides: \(-\frac{1}{50 - m} = 0.002t + C\). Using \(t = 0\), \(m = 0\), we find \(C = -\frac{1}{50}\). Thus, \(-\frac{1}{50 - m} = 0.002t - \frac{1}{50}\). Solving for m, we get \(m = 50 - \frac{500}{t + 10}\).

(iii) Substitute \(t = 10\) into the solution: \(m = 50 - \frac{500}{10 + 10} = 25\). For \(m = 45\), solve \(45 = 50 - \frac{500}{t + 10}\), giving \(t = 90\).

(iv) As \(t\) becomes very large, \(\frac{500}{t + 10}\) approaches 0, so \(m\) approaches 50.