Given \(x = \sin^{-1}\left(\frac{2}{5}\right)\), we have \(\sin x = \frac{2}{5}\).

(i) To find \(\cos^2 x\), use the identity \(\cos^2 x = 1 - \sin^2 x\).

\(\sin^2 x = \left(\frac{2}{5}\right)^2 = \frac{4}{25}\).

Thus, \(\cos^2 x = 1 - \frac{4}{25} = \frac{21}{25}\).

(ii) To find \(\tan^2 x\), use the identity \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\).

\(\tan^2 x = \frac{\frac{4}{25}}{\frac{21}{25}} = \frac{4}{21}\).

(i) We start with the expression \(f(x) = 5 \sin^2 x + 3 \cos^2 x\). Using the identity \(\sin^2 x + \cos^2 x = 1\), we can rewrite \(\cos^2 x\) as \(1 - \sin^2 x\). Thus,

\(f(x) = 5 \sin^2 x + 3(1 - \sin^2 x) = 5 \sin^2 x + 3 - 3 \sin^2 x = 3 + 2 \sin^2 x.\)

Therefore, \(a = 3\) and \(b = 2\).

(ii) We need to solve \(3 + 2 \sin^2 x = 7 \sin x\). Let \(s = \sin x\), then the equation becomes:

\(3 + 2s^2 = 7s.\)

Rearranging gives:

\(2s^2 - 7s + 3 = 0.\)

Using the quadratic formula \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = -7, c = 3\), we find:

\(s = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}.\)

This gives \(s = 3\) or \(s = \frac{1}{2}\). Since \(s = \sin x\) and \(0 \leq x \leq \pi\), \(s = 3\) is not possible. Thus, \(s = \frac{1}{2}\), which gives \(x = \frac{\pi}{6}\) or \(x = \frac{5\pi}{6}\).

(iii) The minimum value of \(f(x) = 3 + 2 \sin^2 x\) occurs when \(\sin^2 x = 0\), giving \(f(x) = 3\). The maximum value occurs when \(\sin^2 x = 1\), giving \(f(x) = 5\). Thus, the range of \(f\) is \(3 \leq f(x) \leq 5\).

(a) The equation \(3 \tan^2 x - 5 \tan x - 2 = 0\) can be factored as \((\tan x - 2)(3 \tan x + 1) = 0\). This gives \(\tan x = 2\) or \(\tan x = -\frac{1}{3}\).

For \(\tan x = 2\), \(x = 63.4^\circ\) (only value in range).

For \(\tan x = -\frac{1}{3}\), \(x = 161.6^\circ\) (only value in range).

(b) For the equation \(3 \tan^2 x - 5 \tan x + k = 0\) to have no solutions, the discriminant must be negative: \(b^2 - 4ac < 0\).

Here, \(b = -5\), \(a = 3\), \(c = k\). So, \((-5)^2 - 4(3)(k) < 0\) simplifies to \(25 - 12k < 0\), leading to \(k > \frac{25}{12}\).

(c) For three solutions, the discriminant must be zero, and \(k = 0\).

Substitute \(k = 0\) into the equation: \(3 \tan^2 x - 5 \tan x = 0\).

Factor to get \(\tan x (3 \tan x - 5) = 0\), giving \(\tan x = 0\) or \(\tan x = \frac{5}{3}\).

For \(\tan x = 0\), \(x = 0^\circ\) or \(x = 180^\circ\).

For \(\tan x = \frac{5}{3}\), \(x = 59.0^\circ\).

(i) We start with the function \(f(x) = 3 \cos^2 x - 2 \sin^2 x\).

Using the identity \(\cos^2 x + \sin^2 x = 1\), we can express \(\sin^2 x\) as \(1 - \cos^2 x\).

Substitute \(\sin^2 x = 1 - \cos^2 x\) into the function:

\(f(x) = 3 \cos^2 x - 2(1 - \cos^2 x)\)

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

\(= 5 \cos^2 x - 2\)

Thus, \(a = 5\) and \(b = -2\).

(ii) To find the range of \(f\), note that \(\cos^2 x\) ranges from 0 to 1 for \(0 \leq x \leq \pi\).

Substitute the extreme values of \(\cos^2 x\) into \(f(x) = 5 \cos^2 x - 2\):

When \(\cos^2 x = 0\), \(f(x) = 5(0) - 2 = -2\).

When \(\cos^2 x = 1\), \(f(x) = 5(1) - 2 = 3\).

Therefore, the range of \(f\) is \([-2, 3]\).

(i) The function \(f(x) = p \sin^2 2x + q\) has a range determined by the values of \(\sin^2 2x\), which vary from 0 to 1. Therefore, the minimum value of \(f(x)\) is \(q\) and the maximum value is \(p + q\). Thus, the range is \(q \leq f(x) \leq p + q\).

(ii) (a) For \(f(x) = p + q\), \(\sin^2 2x = 1\). This occurs at two points within the interval \(0 \leq x \leq \pi\), so there are 2 solutions.

(b) For \(f(x) = q\), \(\sin^2 2x = 0\). This occurs at three points within the interval \(0 \leq x \leq \pi\), so there are 3 solutions.

(c) For \(f(x) = \frac{1}{2}p + q\), \(\sin^2 2x = \frac{1}{2}\). This occurs at four points within the interval \(0 \leq x \leq \pi\), so there are 4 solutions.

(iii) Given \(p = 3\) and \(q = 2\), solve \(3 \sin^2 2x + 2 = 4\) which simplifies to \(\sin^2 2x = \frac{2}{3}\).

\(\sin 2x = \pm \sqrt{\frac{2}{3}}\). Solving for \(2x\), we find \(2x = \sin^{-1}(\pm \sqrt{\frac{2}{3}})\).

Using a calculator, \(2x \approx 0.955, 2.18, 4.09, 5.32\).

Thus, \(x \approx 0.478, 1.09, 2.05, 2.66\).