(a) Given the terms of the geometric progression: \(a = \sin \theta\), \(ar = \cos \theta\), \(ar^2 = 2 - \sin \theta\).

From \(\frac{\cos \theta}{\sin \theta} = \frac{2 - \sin \theta}{\cos \theta}\), we have:

\(\cos^2 \theta = \sin \theta (2 - \sin \theta)\).

Using \(\cos^2 \theta + \sin^2 \theta = 1\), substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) into the equation:

\(1 - \sin^2 \theta = \sin \theta (2 - \sin \theta)\).

Simplifying gives \(\sin^2 \theta + \sin \theta - 1 = 0\).

Solving this quadratic equation for \(\sin \theta\), we find \(\sin \theta = \frac{1}{2}\).

Thus, \(\theta = \frac{\pi}{6}\) or 0.524 radians.

(b) With \(\theta = \frac{\pi}{6}\), \(a = \sin \frac{\pi}{6} = \frac{1}{2}\) and \(r = \sqrt{3}\).

The sum of the first 10 terms is given by:

\(S_{10} = a \frac{1 - r^{10}}{1 - r}\).

Substitute \(a = \frac{1}{2}\) and \(r = \sqrt{3}\):

\(S_{10} = \frac{1}{2} \left( \frac{1 - (\sqrt{3})^{10}}{1 - \sqrt{3}} \right)\).

Calculate \((\sqrt{3})^{10}\) and simplify:

\(S_{10} = \frac{121}{\sqrt{3} - 1}\).

(a)(i) The sum to infinity of a geometric progression is given by \(\frac{a}{1-r} = \frac{1}{\cos \theta}\), where \(a = \cos \theta\) and \(r\) is the common ratio.

Rearranging gives \(1 - r = \cos^2 \theta\), so \(r = 1 - \cos^2 \theta = \sin^2 \theta\).

Thus, the second term is \(a \cdot r = \cos \theta \cdot \sin^2 \theta\).

(a)(ii) The sum of the first 12 terms is given by:

\(S_{12} = \frac{a(1-r^{12})}{1-r}\)

Substituting \(a = \cos \frac{\pi}{3} = 0.5\) and \(r = \sin^2 \frac{\pi}{3} = 0.75\), we have:

\(S_{12} = \frac{0.5(1-(0.75)^{12})}{1-0.75}\)

Calculating gives \(S_{12} = 1.937\).

(b) For the arithmetic progression, the common difference \(d = \cos \theta \sin^2 \theta - \cos \theta\).

Substituting \(\theta = \frac{1}{3} \pi\), we find \(d = -\frac{1}{8}\).

The 85th term is given by:

\(a + 84d = \frac{1}{2} + 84 \left(-\frac{1}{8}\right)\)

Calculating gives the 85th term as \(-10\).

The first term \(a = \sin^2 \theta\).

The second term \(ar = \sin^2 \theta \cos^2 \theta\).

Thus, the common ratio \(r = \cos^2 \theta\).

The sum to infinity of a geometric progression is given by:

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

Substitute \(a = \sin^2 \theta\) and \(r = \cos^2 \theta\):

\(S_\infty = \frac{\sin^2 \theta}{1 - \cos^2 \theta}\)

Since \(1 - \cos^2 \theta = \sin^2 \theta\), we have:

\(S_\infty = \frac{\sin^2 \theta}{\sin^2 \theta} = 1\)

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