(i) Substitute \(x = 5e^{-3t}\) into the differential equation \(\frac{dy}{dt} = -0.6xy\) to get \(\frac{dy}{dt} = -0.6(5e^{-3t})y = -3e^{-3t}y\).

Separate variables: \(\frac{1}{y} dy = -3e^{-3t} dt\).

Integrate both sides: \(\ln y = e^{-3t} + C\).

Use the initial condition \(t = 0, y = 70\) to find \(C\): \(\ln 70 = 1 + C\), so \(C = \ln 70 - 1\).

Thus, \(\ln y = e^{-3t} + \ln 70 - 1\).

Rearrange to find \(y\): \(y = 70(e^{-3t} - 1)\).

(ii) As \(t \to \infty\), \(e^{-3t} \to 0\), so \(y \to 70(-1) = -70\).

The percentage of the initial mass of B remaining is \(p = \frac{y}{70} \times 100\).

Thus, \(p \to \frac{100}{e}\) as \(t \to \infty\).