(i) We are given \(\frac{dN}{dt} = -10\) when \(t = 0\) and \(N = 1000\). The differential equation is \(\frac{dN}{dt} = ke^{-0.02t}N\). Substituting the given values, \(-10 = k \times 1 \times 1000\), we find \(k = -0.01\).

(ii) Separate variables: \(\frac{1}{N} dN = -0.01 e^{-0.02t} dt\). Integrate both sides: \(\int \frac{1}{N} dN = \int -0.01 e^{-0.02t} dt\). This gives \(\ln N = 0.5 e^{-0.02t} + C\). Using \(N = 1000\) when \(t = 0\), \(\ln 1000 = 0.5 + C\), so \(C = \ln 1000 - 0.5\). Substitute \(N = 800\) to find \(t\): \(\ln 800 = 0.5 e^{-0.02t} + \ln 1000 - 0.5\). Solving gives \(t = 29.6\).

(iii) As \(t \to \infty\), \(e^{-0.02t} \to 0\), so \(N \to \frac{1000}{\sqrt{e}}\).