(i) Separate variables: \(\frac{1}{x(4-x)} \frac{dx}{dt} = k\).

Integrate both sides: \(\int \frac{1}{x(4-x)} \, dx = \int k \, dt\).

Use partial fractions: \(\frac{1}{x(4-x)} = \frac{A}{x} + \frac{B}{4-x}\).

Find \(A = \frac{1}{4}\) and \(B = \frac{1}{4}\).

Integrate: \(\frac{1}{4} \ln x - \frac{1}{4} \ln(4-x) = kt + C\).

Use initial conditions \(t = 0, x = 0.4\) and \(t = 2, x = 2\) to find \(C\) and \(k\).

\(\ln \left( \frac{x}{4-x} \right) = 4kt - 8k\).

Substitute \(t = 2, x = 2\): \(\ln 1 = 8k - 8k\).

Substitute \(t = 0, x = 0.4\): \(\ln \left( \frac{0.4}{3.6} \right) = -8k\).

Solve for \(k\): \(k = \frac{1}{4} \ln 3\).

(ii) When 90% of the area is infected, \(x = 3.6\).

Substitute \(x = 3.6\) into \(\ln \left( \frac{x}{4-x} \right) = 4kt - 8k\).

Solve for \(t\): \(t = 4\).