(i) Separate variables: \(\frac{dx}{(20-x)(40-x)} = \frac{1}{5} dt\).

Use partial fractions: \(\frac{1}{(20-x)(40-x)} = \frac{A}{20-x} + \frac{B}{40-x}\).

Find \(A\) and \(B\): \(A = \frac{1}{20}\), \(B = -\frac{1}{20}\).

Integrate both sides: \(\int \left( \frac{1}{20} \ln(20-x) - \frac{1}{20} \ln(40-x) \right) dx = \int \frac{1}{5} dt\).

Integrate: \(-\frac{1}{20} \ln(20-x) + \frac{1}{20} \ln(40-x) = \frac{1}{5} t + C\).

Use initial condition \(x = 10\) when \(t = 0\) to find \(C\).

Solve for \(x\): \(x = \frac{60e^{4t} - 40}{3e^{4t} - 1}\).

(ii) As \(t\) becomes large, \(x\) approaches 20.