(i) To find \(BX\), use the cosine of angle \(30^\circ\) in triangle \(BCX\):

\(BX = BC \cdot \cos(30^\circ) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \text{ cm}\).

To find \(\angle CAB\), use the tangent in triangle \(ACX\):

\(\tan(\angle CAB) = \frac{CX}{AX} = \frac{3}{4 + 3\sqrt{3}}\).

Thus, \(\angle CAB = \tan^{-1} \left( \frac{3}{4 + 3\sqrt{3}} \right)\).

(ii) To find \(AC\), use the Pythagorean theorem in triangle \(ACX\):

\(AC^2 = AX^2 + CX^2 = (4 + 3\sqrt{3})^2 + 3^2\).

\(AC^2 = (16 + 24\sqrt{3} + 27) = 52 + 24\sqrt{3}\).

Thus, \(AC = \sqrt{52 + 24\sqrt{3}} \text{ cm}\).