(a) To solve the differential equation \(e^{3t} \frac{dx}{dt} = \cos^2 2x\), we first separate variables:

\(\int \sec^2 2x \, dx = \int e^{-3t} \, dt\).

Integrating both sides, we get:

\(\frac{1}{2} \tan 2x = -\frac{1}{3} e^{-3t} + C\).

Using the initial condition \(x = 0\) when \(t = 0\), we find \(C = \frac{1}{3}\).

Thus, \(\frac{1}{2} \tan 2x = -\frac{1}{3} e^{-3t} + \frac{1}{3}\).

Solving for \(x\), we obtain:

\(x = \frac{1}{2} \arctan \left( \frac{2}{3} (1 - e^{-3t}) \right)\).

(b) As \(t \to \infty\), \(e^{-3t} \to 0\), so \(x \to \frac{1}{2} \arctan \left( \frac{2}{3} \right)\).