(i) To find AC, use the cosine of angle DAC: \(AC = l\cos 30^\circ = \frac{l\sqrt{3}}{2}\).

To find BC, use the sine of angle DAC: \(BC = 2l\sin 30^\circ = l\).

To find AB, use the Pythagorean theorem in triangle ABD: \(AB = \sqrt{AD^2 + BD^2} = \sqrt{l^2 + \left(\frac{l}{2}\right)^2} = \frac{1}{2}l\sqrt{7}\).

(ii) Use the tangent of angle \((x + 30^\circ)\) in triangle ABC: \(\tan(x + 30^\circ) = \frac{BC}{AC} = \frac{l}{\frac{l\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}\).

Therefore, \(x = \tan^{-1}\!\left(\frac{2}{\sqrt{3}}\right) - 30^\circ\).

(i) Using the tangent ratio in triangle ABC, \(\tan \left( \frac{\pi}{3} \right) = \frac{AC}{2x}\). Therefore, \(AC = 2x \sqrt{3}\). Using the cosine ratio, \(\cos \left( \frac{\pi}{3} \right) = \frac{2x}{AB}\), we find \(AB = 4x\). Using the Pythagorean theorem in triangle ABM, \(AM = \sqrt{AB^2 - BM^2} = \sqrt{(4x)^2 - x^2} = \sqrt{16x^2 - x^2} = \sqrt{15x^2} = \sqrt{15}x\).

(ii) In triangle AMC, \(\tan(\angle MAC) = \frac{x}{AC} = \frac{x}{2x \sqrt{3}} = \frac{1}{2 \sqrt{3}}\). Therefore, \(\angle MAC = \tan^{-1} \left( \frac{1}{2 \sqrt{3}} \right)\). Since \(\angle BAM = \theta\) and \(\angle BAC = \frac{\pi}{6}\), we have \(\theta = \angle BAC - \angle MAC = \frac{\pi}{6} - \tan^{-1} \left( \frac{1}{2 \sqrt{3}} \right)\).

To find the length of \(BC\), we use the sine rule:

\(\frac{AB}{\sin ACB} = \frac{BC}{\sin BAC}\)

Substitute the known values:

\(\frac{12}{\sin 45°} = \frac{x}{\sin 60°}\)

We know:

\(\sin 60° = \frac{\sqrt{3}}{2}\) and \(\sin 45° = \frac{1}{\sqrt{2}}\).

Substitute these into the equation:

\(\frac{12}{\frac{1}{\sqrt{2}}} = \frac{x}{\frac{\sqrt{3}}{2}}\)

Cross-multiply to solve for \(x\):

\(12 \times \frac{\sqrt{3}}{2} = x \times \frac{1}{\sqrt{2}}\)

\(12 \times \frac{\sqrt{3}}{2} \times \sqrt{2} = x\)

\(x = 6\sqrt{6}\)

Thus, the exact length of \(BC\) is \(6\sqrt{6}\).

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

(i) To find the length of CD, use the sine rule in triangle ABC:

\(CD = AB \sin(30°) + BC \sin(60°)\)

\(= 2d \cdot \frac{1}{2} + (2\sqrt{3})d \cdot \frac{\sqrt{3}}{2}\)

\(= d + 3d = 4d\)

(ii) To show that angle CAD = tan^-1(\frac{2}{\sqrt{3}}), use the tangent rule:

\(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{ans to (i)}}{AB \cos(30°) + BC \cos(60°)}\)

\(= \frac{4d}{2d \cdot \frac{\sqrt{3}}{2} + (2\sqrt{3})d \cdot \frac{1}{2}}\)

\(= \frac{4d}{d\sqrt{3} + d\sqrt{3}}\)

\(= \frac{4d}{2d\sqrt{3}}\)

\(= \frac{2}{\sqrt{3}}\)

Thus, \(\theta = \tan^{-1}\left(\frac{2}{\sqrt{3}}\right)\).

(i) To find AC, use the cosine of angle DAC: \(AC = l\cos 30^\circ = \frac{l\sqrt{3}}{2}\).

To find BC, use the sine of angle DAC: \(BC = 2l\sin 30^\circ = l\).

To find AB, use the Pythagorean theorem in triangle ABD: \(AB = \sqrt{AD^2 + BD^2} = \sqrt{l^2 + \left(\frac{l}{2}\right)^2} = \frac{1}{2}l\sqrt{7}\).

(ii) Use the tangent of angle \((x + 30^\circ)\) in triangle ABC: \(\tan(x + 30^\circ) = \frac{BC}{AC} = \frac{l}{\frac{l\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}\).

Therefore, \(x = \tan^{-1}\!\left(\frac{2}{\sqrt{3}}\right) - 30^\circ\).