(a) The terms of the arithmetic progression are given by:

First term: \(a = 2 \cos x\)

Third term: \(a + 2d = -6\sqrt{3} \sin x\)

Fifth term: \(a + 4d = 10 \cos x\)

From the third and first terms: \(-6\sqrt{3} \sin x - 2 \cos x = 2d\)

From the fifth and third terms: \(10 \cos x + 6\sqrt{3} \sin x = 2d\)

Equating the expressions for \(2d\):

\(-6\sqrt{3} \sin x - 2 \cos x = 10 \cos x + 6\sqrt{3} \sin x\)

\(-12\sqrt{3} \sin x = 12 \cos x\)

\(\tan x = -\frac{1}{\sqrt{3}}\)

\(x = \frac{5\pi}{6}\) (since \(\frac{1}{2}\pi < x < \pi\))

(b) The first term \(a = 2 \cos x = -\sqrt{3}\) and common difference \(d = 2 \cos x = -\sqrt{3}\).

The sum of the first 25 terms is given by:

\(S_{25} = \frac{25}{2} \left(2a + (25-1)d\right)\)

\(S_{25} = \frac{25}{2} \left(2(-\sqrt{3}) + 24(-\sqrt{3})\right)\)

\(S_{25} = \frac{25}{2} \times (-50\sqrt{3})\)

\(S_{25} = -325\sqrt{3}\)

(a) The common difference \(d\) is given by the difference between the second term and the first term:

\(d = -\frac{\tan^2 \theta}{\cos^2 \theta} - \frac{1}{\cos^2 \theta}\)

\(= \frac{-\sin^2 \theta - \cos^2 \theta}{\cos^4 \theta}\)

\(= \frac{-1}{\cos^4 \theta}\)

Thus, the common difference is \(-\frac{1}{\cos^4 \theta}\).

(b) The 13th term \(u_{13}\) is given by:

\(u_{13} = a + 12d\)

Where \(a = \frac{1}{\cos^2 \theta}\) and \(d = -\frac{1}{\cos^4 \theta}\).

When \(\theta = \frac{1}{6} \pi\), \(\cos \theta = \frac{\sqrt{3}}{2}\).

\(a = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{4}{3}\)

\(d = -\frac{1}{\left(\frac{\sqrt{3}}{2}\right)^4} = -\frac{16}{9}\)

\(u_{13} = \frac{4}{3} + 12 \left(-\frac{16}{9}\right)\)

\(= \frac{4}{3} - \frac{192}{9}\)

\(= \frac{4}{3} - \frac{64}{3}\)

\(= -20\)

Thus, the exact value of the 13th term is \(-20\).

(i) The first term \(a = \sin^2 \theta\) and the second term \(a + d = \sin^2 \theta \cos^2 \theta\).

The common difference \(d = \sin^2 \theta \cos^2 \theta - \sin^2 \theta\).

Factor out \(\sin^2 \theta\):

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

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

\(d = -\sin^4 \theta\).

(ii) The sum of the first 16 terms of an arithmetic progression is given by:

\(S_{16} = \frac{16}{2} [2a + 15d]\).

With \(\theta = \frac{1}{3} \pi\), \(\sin \theta = \frac{\sqrt{3}}{2}\), so \(a = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\).

\(d = -\left(\frac{\sqrt{3}}{2}\right)^4 = -\frac{9}{16}\).

Substitute \(a\) and \(d\) into the sum formula:

\(S_{16} = 8 \left[2 \times \frac{3}{4} + 15 \times -\frac{9}{16}\right]\).

\(S_{16} = 8 \left[\frac{3}{2} - \frac{135}{16}\right]\).

\(S_{16} = 8 \left[\frac{24}{16} - \frac{135}{16}\right]\).

\(S_{16} = 8 \times -\frac{111}{16}\).

\(S_{16} = -\frac{888}{16} = -\frac{55}{2}\).

The first term \(a = \cos \theta\) and the second term \(a + d = \cos \theta + \sin^2 \theta\) gives the common difference \(d = \sin^2 \theta\).

The sum of the first 13 terms of an arithmetic progression is given by:

\(S_{13} = \frac{13}{2} \left[ 2a + 12d \right] = 52\)

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

\(\frac{13}{2} \left[ 2\cos \theta + 12\sin^2 \theta \right] = 52\)

Simplify and solve for \(\theta\):

\(2\cos \theta + 12(1 - \cos^2 \theta) = 8\)

\(2\cos \theta + 12 - 12\cos^2 \theta = 8\)

\(12\cos^2 \theta - 2\cos \theta - 4 = 0\)

Divide by 2:

\(6\cos^2 \theta - \cos \theta - 2 = 0\)

Solving the quadratic equation for \(\cos \theta\):

\(\cos \theta = \frac{2}{3} \text{ or } \cos \theta = -\frac{1}{2}\)

For \(\cos \theta = \frac{2}{3}\), \(\theta \approx 0.841\).

For \(\cos \theta = -\frac{1}{2}\), \(\theta \approx 2.09\).

(i) The terms of the arithmetic progression are \(2 \sin x\), \(3 \cos x\), and \(\sin x + 2 \cos x\). The common difference \(d\) is given by:

\(3 \cos x - 2 \sin x = (\sin x + 2 \cos x) - 3 \cos x\)

Simplifying, we have:

\(3 \cos x - 2 \sin x = \sin x - \cos x\)

Rearranging gives:

\(4 \cos x = 3 \sin x\)

Thus, \(\tan x = \frac{3}{4}\).

(ii) Using \(\cos x = \frac{3}{5}\) and \(\sin x = \frac{4}{5}\), we find:

\(a = 2 \sin x = 2 \times \frac{4}{5} = 1.6\)

\(d = 3 \cos x - 2 \sin x = 0.2\)

The sum of the first 20 terms is given by:

\(S_{20} = \frac{20}{2} (2 \times 1.6 + (20 - 1) \times 0.2) = 70\)

The first term \(a_1 = 1\) and the second term \(a_2 = \cos^2 x\). The common difference \(d\) is given by:

\(d = a_2 - a_1 = \cos^2 x - 1\).

The sum of the first ten terms \(S_{10}\) of an arithmetic progression is given by:

\(S_{10} = \frac{10}{2} (2a_1 + 9d)\).

Substitute \(a_1 = 1\) and \(d = \cos^2 x - 1\):

\(S_{10} = 5 [2 \times 1 + 9(\cos^2 x - 1)]\).

Simplify the expression:

\(S_{10} = 5 [2 + 9\cos^2 x - 9]\).

\(S_{10} = 5 [-7 + 9\cos^2 x]\).

Using the identity \(\cos^2 x = 1 - \sin^2 x\), substitute:

\(S_{10} = 5 [-7 + 9(1 - \sin^2 x)]\).

\(S_{10} = 5 [-7 + 9 - 9\sin^2 x]\).

\(S_{10} = 5 [2 - 9\sin^2 x]\).

\(S_{10} = 10 - 45\sin^2 x\).

Thus, \(a = 10\) and \(b = 45\).