Answer: (a)(i) \(\theta=\frac\pi3\), (ii) \(20\sqrt3\); (b)(i) \(\phi=\pm\frac\pi3\), (ii) no sum to infinity.

Use the appropriate formula for an arithmetic or geometric sequence, and check the common difference or common ratio before substituting.

(a) Let

\(t=\tan\frac{\theta}{2}\).

The first three terms are \(-3t\), \(-t\), and \(t\), so the common difference is

\(2t\).

The 12th term is

\(-3t+11(2t)=19t\).

Given that this is \(\frac{19\sqrt3}{3}\),

\(19t=\frac{19\sqrt3}{3}\).

So

\(t=\frac{\sqrt3}{3}\).

Hence

\(\tan\frac{\theta}{2}=\frac{\sqrt3}{3}\).

Since \(0\lt \theta\lt \frac12\pi\), we have \(0\lt \frac\theta2\lt \frac14\pi\). Therefore

\(\frac\theta2=\frac\pi6\),

and so

\(\theta=\frac\pi3\).

The sum of the first 10 terms is

\(S_{10}=\frac{10}{2}\big(2(-3t)+9(2t)\big)\).

So

\(S_{10}=5(-6t+18t)=60t\).

Using \(t=\frac{\sqrt3}{3}\),

\(S_{10}=60\cdot\frac{\sqrt3}{3}=20\sqrt3\).

(b) The common ratio is

\(r=\frac{\frac14\operatorname{cosec}^2\phi}{\frac1{16}\operatorname{cosec}^4\phi}=4\sin^2\phi\).

The third term is \(1\), so the fourth term is \(r\).

Given that the sum of the 3rd and 4th terms is 4,

\(1+r=4\).

So

\(r=3\).

Therefore

\(4\sin^2\phi=3\).

Hence

\(\sin^2\phi=\frac34\),

so

\(\sin\phi=\pm\frac{\sqrt3}{2}\).

Since \(-\frac12\pi\lt \phi\lt \frac12\pi\), the possible values are

\(\phi=\frac\pi3\) and \(\phi=-\frac\pi3\).

For a geometric progression to have a sum to infinity, we need \(|r|\lt 1\).

Here \(r=3\), so \(|r|\gt 1\).

Therefore this geometric progression does not have a sum to infinity.