Answer: (a) \(r=3\), first term \(=\frac23\). (b) \(-90^\circ\lt \theta\lt -60^\circ\) or \(60^\circ\lt \theta\lt 90^\circ\).

Use the standard trigonometric identities and make sure the final angles are chosen from the interval given in the question.

Let the first term be \(a\) and the common ratio be \(r\).

The 3rd term is \(ar^2=6\), and the 8th term is \(ar^7=1458\).

Dividing the second equation by the first gives

\(r^5=\frac{1458}{6}=243\).

So \(r=3\).

Then \(a\cdot3^2=6\), giving \(a=\frac23\).

For the second geometric progression, the common ratio is

\(r=\frac{2\cos^2\theta}{\cos\theta}=2\cos\theta\).

A geometric progression has a sum to infinity when \(|r|\lt 1\), so

\(|2\cos\theta|\lt 1\).

Thus \(-\frac12\lt \cos\theta\lt \frac12\).

In the interval \(-90^\circ\lt \theta\lt 90^\circ\), this gives

\(-90^\circ\lt \theta\lt -60^\circ\) or \(60^\circ\lt \theta\lt 90^\circ\).

This gives the required answer: (a) \(r=3\), first term \(=\frac23\). (b) \(-90^\circ\lt \theta\lt -60^\circ\) or \(60^\circ\lt \theta\lt 90^\circ\).