(i) To find an expression for \(N\) in terms of \(t\), we start by separating variables and integrating both sides:

\(\int \frac{1}{N^{0.5}} \, dN = \int 1.2e^{-0.02t} \, dt\).

This gives:

\(2N^{0.5} = -60e^{-0.02t} + C\).

Using the initial condition \(N = 100\) when \(t = 0\), we find \(C\):

\(2(100)^{0.5} = -60e^{0} + C\)

\(20 = -60 + C\)

\(C = 80\).

Thus, the equation becomes:

\(2N^{0.5} = -60e^{-0.02t} + 80\).

Solving for \(N\):

\(N^{0.5} = -30e^{-0.02t} + 40\)

\(N = (40 - 30e^{-0.02t})^2\).

(ii) As \(t \to \infty\), \(e^{-0.02t} \to 0\), so \(N \to 40^2 = 1600\).