(i) Separate variables: \(\frac{dx}{\cos^2 x} = e^{-2t} dt\).

Integrate both sides: \(\int \sec^2 x \, dx = \int e^{-2t} \, dt\).

This gives \(\tan x = -\frac{1}{2} e^{-2t} + C\).

Use initial condition \(t = 0, x = 0\): \(\tan 0 = 0 = -\frac{1}{2} e^{0} + C\) so \(C = \frac{1}{2}\).

Thus, \(\tan x = \frac{1}{2} - \frac{1}{2} e^{-2t}\).

Rearrange to find \(x\): \(x = \arctan\left(\frac{1}{2} - \frac{1}{2} e^{-2t}\right)\).

(ii) As \(t \to \infty\), \(e^{-2t} \to 0\), so \(x \to \arctan\left(\frac{1}{2}\right)\).

(iii) As \(t\) increases, \(e^{-2t}\) decreases, making \(1 - e^{-2t}\) increase, and thus \(x\) increases.