(i) To find the stationary points, we first find the derivative of \(y = 3 \sin x + 4 \cos^3 x\).

The derivative is \(\frac{dy}{dx} = 3 \cos x - 12 \cos^2 x \sin x\).

Set \(\frac{dy}{dx} = 0\):

\(3 \cos x - 12 \cos^2 x \sin x = 0\)

Factor out \(\cos x\):

\(\cos x (3 - 12 \cos x \sin x) = 0\)

This gives \(\cos x = 0\) or \(3 - 12 \cos x \sin x = 0\).

For \(\cos x = 0\), \(x = \frac{1}{2}\pi\).

For \(3 - 12 \cos x \sin x = 0\), solve for \(\sin 2x = \frac{1}{2}\):

\(2x = \frac{\pi}{6}, \frac{5\pi}{6}\)

Thus, \(x = \frac{1}{12}\pi, \frac{5}{12}\pi\).

Therefore, the \(x\)-coordinates are \(x = \frac{1}{12}\pi, \frac{5}{12}\pi, \frac{1}{2}\pi\).

(ii) To determine the nature of the stationary point at \(x = \frac{1}{12}\pi\), we check the second derivative or use the first derivative test.

Since the mark scheme indicates a maximum at \(x = \frac{1}{12}\pi\), this is a maximum point.