(i) Given \(\frac{dy}{dt} = -0.2xy\) and \(x = \frac{10}{(1+t)^2}\), substitute for x:

\(\frac{dy}{dt} = -0.2 \cdot \frac{10}{(1+t)^2} \cdot y = -\frac{2y}{(1+t)^2}\).

Separate variables: \(\frac{dy}{y} = -\frac{2}{(1+t)^2} dt\).

Integrate both sides: \(\int \frac{dy}{y} = \int -\frac{2}{(1+t)^2} dt\).

\(\ln y = \frac{2}{1+t} + C\).

Use initial condition \(y = 100\) when \(t = 0\):

\(\ln 100 = 2 + C\).

Thus, \(C = \ln 100 - 2\).

Final solution: \(\ln y = \frac{2}{1+t} - 2 + \ln 100\).

(ii) As \(t \to \infty\), \(\frac{2}{1+t} \to 0\), so \(\ln y \to \ln 100 - 2\).

Thus, \(y \to e^{\ln 100 - 2} = \frac{100}{e^2}\).

The mass of A, \(x = \frac{10}{(1+t)^2}\), tends to zero as \(t \to \infty\).