(i) Start with the identity \(\cos^2 x + \sin^2 x = 1\). Raise both sides to the power of 3:

\((\cos^2 x + \sin^2 x)^3 = 1^3 = 1\).

Expand \((\cos^2 x + \sin^2 x)^3\):

\((\cos^2 x + \sin^2 x)^3 = \cos^6 x + 3\cos^4 x \sin^2 x + 3\cos^2 x \sin^4 x + \sin^6 x\).

Using \(\cos^2 x + \sin^2 x = 1\), substitute \(\cos^2 x = 1 - \sin^2 x\) and \(\sin^2 x = 1 - \cos^2 x\):

\(\cos^6 x + \sin^6 x = 1 - 3\cos^2 x \sin^2 x\).

Using the double angle identity \(\sin 2x = 2 \sin x \cos x\), we have \(\sin^2 2x = 4 \sin^2 x \cos^2 x\):

\(\cos^6 x + \sin^6 x = 1 - \frac{3}{4} \sin^2 2x\).

(ii) Given \(\cos^6 x + \sin^6 x = \frac{2}{3}\), substitute the result from part (i):

\(1 - \frac{3}{4} \sin^2 2x = \frac{2}{3}\).

Solve for \(\sin^2 2x\):

\(\frac{3}{4} \sin^2 2x = 1 - \frac{2}{3} = \frac{1}{3}\).

\(\sin^2 2x = \frac{4}{9}\).

\(\sin 2x = \pm \frac{2}{3}\).

For \(\sin 2x = \frac{2}{3}\), solve for \(2x\):

\(2x = \sin^{-1}\left(\frac{2}{3}\right)\) gives \(2x \approx 41.8^\circ\) or \(2x \approx 138.2^\circ\).

Thus, \(x \approx 20.9^\circ\) or \(x \approx 69.1^\circ\).

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

\(2x = \sin^{-1}\left(-\frac{2}{3}\right)\) gives \(2x \approx 221.8^\circ\) or \(2x \approx 318.2^\circ\).

Thus, \(x \approx 110.9^\circ\) or \(x \approx 159.1^\circ\).

Therefore, the solutions are \(x = 20.9^\circ, 69.1^\circ, 110.9^\circ, 159.1^\circ\).