(i) Express \(\frac{1}{y(4-y)}\) in partial fractions: \(\frac{A}{y} + \frac{B}{4-y}\).

Solving for \(A\) and \(B\), we get \(A = \frac{1}{4}\) and \(B = -\frac{1}{4}\).

Thus, \(\frac{1}{y(4-y)} = \frac{1}{4} \left( \frac{1}{y} - \frac{1}{4-y} \right)\).

Integrate: \(\int \frac{1}{4} \left( \frac{1}{y} - \frac{1}{4-y} \right) \, dy = \frac{1}{4} \ln y - \frac{1}{4} \ln (4-y) + C\).

(ii) Separate variables: \(\frac{dy}{y(4-y)} = dx\).

Integrate both sides: \(\int \left( \frac{1}{y} + \frac{1}{4-y} \right) \, dy = \int dx\).

Using initial condition \(y = 1\) when \(x = 0\), solve for constant.

Final expression: \(y = \frac{4}{3e^{-4x} + 1}\).

(iii) As \(x \to \infty\), \(e^{-4x} \to 0\), so \(y \to 4\).