Answer: (a) \(n=57\); (b) \(p^{(5-2n)x}\); (c) \(\frac{\pi}{18}\lt\theta\lt\frac{5\pi}{18}\).

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

(a) The first three terms are

\(\ln q,\quad \ln q^4,\quad \ln q^7.\)

Since \(\ln q^4=4\ln q\) and \(\ln q^7=7\ln q\), the common difference is \(3\ln q\).

The sum of the first \(n\) terms is

\(S_n=\frac n2\left(2\ln q+(n-1)3\ln q\right).\)

So

\(S_n=\frac n2(3n-1)\ln q.\)

Given \(S_n=4845\ln q\),

\(\frac n2(3n-1)=4845.\)

Hence

\(n(3n-1)=9690.\)

So

\(3n^2-n-9690=0.\)

Factorising gives the positive solution

\(n=57.\)

(b) The first term is \(p^{3x}\). The common ratio is

\(\frac{p^x}{p^{3x}}=p^{-2x}.\)

Therefore the \(n\)th term is

\(p^{3x}\left(p^{-2x}\right)^{n-1}.\)

Simplify the exponent:

\(3x-2x(n-1)=(5-2n)x.\)

So the \(n\)th term is

\(p^{(5-2n)x}.\)

(c) The common ratio is

\(r=\frac{\frac{16}{9}\cos^4 3\theta}{\frac43\cos^2 3\theta} =\frac43\cos^2 3\theta.\)

For a sum to infinity,

\(\lvert r\rvert\lt1.\)

Since \(\cos^2 3\theta\geq0\), this becomes

\(\frac43\cos^2 3\theta\lt1.\)

So

\(\cos^2 3\theta\lt\frac34.\)

Given \(0\lt\theta\lt\frac{\pi}{3}\), we have

\(0\lt3\theta\lt\pi.\)

In this interval, \(\cos^2 3\theta\lt\frac34\) gives

\(\frac{\pi}{6}\lt3\theta\lt\frac{5\pi}{6}.\)

Dividing by 3,

\(\frac{\pi}{18}\lt\theta\lt\frac{5\pi}{18}.\)