For a geometric progression to be convergent, the common ratio \(r\) must satisfy \(-1 < r < 1\).

The common ratio \(r\) is given by:

\(r = \frac{2 \cos \theta}{\sqrt{3}}\)

Thus, the inequality becomes:

\(-1 < \frac{2 \cos \theta}{\sqrt{3}} < 1\)

We solve the inequality \(0 < \frac{2 \cos \theta}{\sqrt{3}} < 1\) since \(0 < \theta < \pi\):

\(0 < 2 \cos \theta < \sqrt{3}\)

\(0 < \cos \theta < \frac{\sqrt{3}}{2}\)

The values of \(\theta\) for which \(\cos \theta = \frac{\sqrt{3}}{2}\) are \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

Therefore, the set of values of \(\theta\) for which the progression is convergent is:

\(\frac{\pi}{6} < \theta < \frac{5\pi}{6}\).

(i) For a geometric progression to be convergent, the common ratio \(r\) must satisfy \(|r| < 1\). Here, \(r = \frac{1}{3} \tan^2 \theta\). Therefore, we need:

\(0 < \frac{1}{3} \tan^2 \theta < 1\)

This simplifies to:

\(0 < \tan^2 \theta < 3\)

Taking the square root, we get:

\(0 < \tan \theta < \sqrt{3}\)

Since \(\tan \theta = \sqrt{3}\) when \(\theta = \frac{\pi}{3}\), the set of values for \(\theta\) is:

\(0 < \theta < \frac{\pi}{3}\)

(ii) The sum to infinity \(S_\infty\) of a geometric progression is given by:

\(S_\infty = \frac{a}{1 - r}\)

where \(a = 1\) and \(r = \frac{1}{3} \tan^2 \frac{\pi}{6}\).

Since \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\), we have:

\(\tan^2 \frac{\pi}{6} = \frac{1}{3}\)

Thus, \(r = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\).

Therefore,

\(S_\infty = \frac{1}{1 - \frac{1}{9}} = \frac{1}{\frac{8}{9}} = \frac{9}{8}\)

So, \(S_\infty = \frac{9}{8} \text{ or } 1.125\).