(i) We have \(gf(x) = 1\), which means \(g(f(x)) = 1\). Therefore, \(\cos\left(\frac{1}{2}x + \frac{\pi}{6}\right) = 1\).

The cosine function equals 1 at \(0\), so \(\frac{1}{2}x + \frac{\pi}{6} = 0\).

Solving for \(x\), we get:

\(\frac{1}{2}x = -\frac{\pi}{6}\)

\(x = -\frac{\pi}{3}\).

(ii) We have \(fg(x) = 1\), which means \(f(g(x)) = 1\). Therefore, \(\frac{1}{2}\cos x + \frac{\pi}{6} = 1\).

Solving for \(\cos x\), we get:

\(\frac{1}{2}\cos x = 1 - \frac{\pi}{6}\)

\(\cos x = 2\left(1 - \frac{\pi}{6}\right)\)

\(\cos x = 0.9528\)

Solving for \(x\), we find \(x = \pm 0.31\) (to 2 decimal places).