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