First, evaluate \(-\cos^{-1}\left(\frac{1}{2}\sqrt{3}\right)\). The value of \(\cos^{-1}\left(\frac{1}{2}\sqrt{3}\right)\) is \(\frac{\pi}{6}\), so \(-\cos^{-1}\left(\frac{1}{2}\sqrt{3}\right) = -\frac{\pi}{6}\).

Substitute this into the equation:

\(\frac{1}{6}\pi + \arctan(4x) = -\frac{\pi}{6}\).

Rearrange to find \(\arctan(4x)\):

\(\arctan(4x) = -\frac{\pi}{6} - \frac{1}{6}\pi = -\frac{\pi}{3}\).

Thus, \(4x = \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}\).

Solving for \(x\), we get:

\(x = -\frac{\sqrt{3}}{4}\).

(i) To find the \(x\)-coordinate of \(A\), set \(\sin x = 2 \cos x\).

This implies \(\tan x = 2\).

Solving for \(x\), we find \(x = 1.11\) (or approximately \(0.352\pi\)).

(ii) For the \(y\)-coordinate of \(B\), we need the value of \(y\) when \(x\) is such that \(y = \sin x = 2 \cos x\) and \(y < 0\).

The negative solution for \(y\) is in the range \(-1 < y < -0.8\).

The value is approximately \(y = -0.895\).

(i) We have the equations:

\(a + \frac{1}{2}b = 5\)

\(a - b = 11\)

Solving these simultaneously:

From \(a - b = 11\), we get \(a = 11 + b\).

Substitute into \(a + \frac{1}{2}b = 5\):

\(11 + b + \frac{1}{2}b = 5\)

\(11 + \frac{3}{2}b = 5\)

\(\frac{3}{2}b = -6\)

\(b = -4\)

Substitute \(b = -4\) into \(a = 11 + b\):

\(a = 11 - 4 = 7\)

Thus, \(a = 7\) and \(b = -4\).

(ii) The function \(f(x) = a + b \cos x\) has a range determined by the maximum and minimum values of \(\cos x\), which are 1 and -1, respectively.

Thus, the range of \(f(x)\) is:

\(a + b \leq f(x) \leq a - b\)

Substitute \(a = 7\) and \(b = -4\):

\(7 - 4 \leq f(x) \leq 7 + 4\)

\(3 \leq f(x) \leq 11\)

For \(f(x) = k\) to have no solution, \(k\) must be outside this range:

\(k < 3\) or \(k > 11\).

The maximum value of the sine function is 1, so at \(\theta = 0^\circ\), \(y = k \sin(\alpha) = 2\). Therefore, \(k \sin(\alpha) = 2\).

At \(\theta = 150^\circ\), the graph crosses the \(\theta\)-axis, so \(y = k \sin(150^\circ + \alpha) = 0\). This implies \(150^\circ + \alpha = 180^\circ\), so \(\alpha = 30^\circ\).

Substituting \(\alpha = 30^\circ\) into \(k \sin(\alpha) = 2\), we have \(k \sin(30^\circ) = 2\).

Since \(\sin(30^\circ) = \frac{1}{2}\), we get \(k \cdot \frac{1}{2} = 2\).

Therefore, \(k = 4\).

(i) The line cuts the x-axis at A, so the y-coordinate of A is 0. The angle \(\angle BAO = \frac{1}{6} \pi\) radians implies that the slope of the line is \(\tan\left(\frac{1}{6} \pi\right) = \frac{1}{\sqrt{3}}\).

The equation of the line is \(y - 2 = \frac{1}{\sqrt{3}}(x - 0)\).

Setting \(y = 0\) to find the x-coordinate of A:

\(0 - 2 = \frac{1}{\sqrt{3}}x\)

\(x = 2\sqrt{3}\)

(ii) The midpoint of AB is \(\left(\frac{\sqrt{3}}{2}, 1\right)\).

The gradient of AB is \(-\frac{1}{\sqrt{3}}\), so the gradient of the perpendicular bisector is \(\sqrt{3}\).

The equation of the perpendicular bisector is \(y - 1 = \sqrt{3}(x - \frac{\sqrt{3}}{2})\).

Simplifying gives \(y = \sqrt{3}x - 2\).