(i) For the curve \(y = 3 \cos 2x\), the maximum value of \(\cos 2x\) is 1 and the minimum is -1. Therefore, the largest value of \(y\) is 3 and the smallest is -3.

For the line \(2y + \frac{3x}{\pi} = 5\), rearrange to find \(y = \frac{5}{2} - \frac{3x}{2\pi}\). The smallest value of \(y\) occurs when \(x = 2\pi\), giving \(y = -\frac{1}{2}\), and the largest value occurs when \(x = 0\), giving \(y = \frac{5}{2}\).

(ii) The sketch should show two complete oscillations of the cosine curve starting with a maximum at \((0, 3)\) and leveling off at \(0\) and \(2\pi\). The line should start on the positive \(y\)-axis and finish below the \(x\)-axis at \(2\pi\).

(iii) The number of solutions of the equation \(6 \cos 2x = 5 - \frac{3x}{\pi}\) for \(0 \leq x \leq 2\pi\) is 4.

(i) We know that \(\sin^2 x + \cos^2 x = 1\).

Substitute \(\sin x = a + b\) and \(\cos x = a - b\):

\((a + b)^2 + (a - b)^2 = 1\)

Expanding both terms, we get:

\(a^2 + 2ab + b^2 + a^2 - 2ab + b^2 = 1\)

Combine like terms:

\(2a^2 + 2b^2 = 1\)

Divide by 2:

\(a^2 + b^2 = \frac{1}{2}\)

(ii) Given \(\tan x = 2\), we have:

\(\tan x = \frac{\sin x}{\cos x} = \frac{a + b}{a - b} = 2\)

Cross-multiply to solve for a:

\(a + b = 2(a - b)\)

\(a + b = 2a - 2b\)

Rearrange terms:

\(b + 2b = 2a - a\)

\(3b = a\)

Thus, \(a = 3b\).

Given the equation \(y = a + \tan bx\), we know the curve intersects the \(y\)-axis at \((0, \sqrt{3})\). Substituting \(x = 0\) into the equation gives:

\(\sqrt{3} = a + \tan(0)\)

\(\sqrt{3} = a\)

Thus, \(a = \sqrt{3}\).

Next, the curve intersects the \(x\)-axis at \(\left(-\frac{1}{6}\pi, 0\right)\). Substituting \(y = 0\) and \(x = -\frac{1}{6}\pi\) into the equation gives:

\(0 = \sqrt{3} + \tan\left(-\frac{b}{6}\pi\right)\)

\(\tan\left(-\frac{b}{6}\pi\right) = -\sqrt{3}\)

The angle whose tangent is \(-\sqrt{3}\) is \(-\frac{\pi}{3}\), so:

\(-\frac{b}{6}\pi = -\frac{\pi}{3}\)

Solving for \(b\):

\(\frac{b}{6} = \frac{1}{3}\)

\(b = 2\)

Thus, \(b = 2\).