To find the length of AX, we use the tangent of the angle \(\frac{1}{6} \pi\) (half of \(\frac{1}{3} \pi\)) in the right triangle formed by O, A, and X.

\(\tan \frac{1}{6} \pi = \frac{AX}{12}\)

\(\tan \frac{1}{6} \pi = \frac{\sqrt{3}}{3}\)

Thus, \(\frac{\sqrt{3}}{3} = \frac{AX}{12}\)

Solving for \(AX\), we get:

\(AX = 12 \times \frac{\sqrt{3}}{3} = 4\sqrt{3}\)

First, calculate the length of OC and AC using trigonometry. Since \(\angle AOB = \frac{1}{3} \pi\), we have:

\(OC = 12 \cos\left(\frac{1}{3} \pi\right) = 12 \times \frac{1}{2} = 6\)

\(AC = 12 \sin\left(\frac{1}{3} \pi\right) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}\)

The area of the sector AOB is given by:

\(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 12^2 \times \frac{1}{3} \pi = 24\pi\)

The area of rectangle OCDB is:

\(12 \times 6\sqrt{3} = 72\sqrt{3}\)

The area of triangle OAC is:

\(\frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\)

The area of the shaded region is the area of the rectangle minus the area of the sector and the triangle:

\(72\sqrt{3} - (24\pi + 18\sqrt{3}) = 54\sqrt{3} - 24\pi\)

Thus, the values of a and b are 54 and 24, respectively.

(i) To find angle AOB, use the tangent formula in the right triangle formed by the radius and the tangent: \(\tan(\frac{x}{2}) = \frac{15}{8} = 1.875\). Solving for \(\frac{x}{2}\), we get \(\frac{x}{2} = 1.081\). Therefore, \(x = 2.16\) radians.

(ii) The perimeter of the shaded region is the sum of the lengths of AT, BT, and the arc AB. The arc length is given by \(r\theta = 8 \times 2.16 = 17.3\) cm. Thus, the perimeter is \(15 + 15 + 17.3 = 47.3\) cm.

(iii) The area of sector AOB is \(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 8^2 \times 2.16 = 69.1\) cm². The area of triangle AOT is \(2 \times \frac{1}{2} \times 8 \times 15 = 120\) cm². The shaded area is \(120 - 69.1 = 50.9\) cm².

The gradient of the line \(l\) is found by rearranging \(2x + y = k\) to \(y = -2x + k\). Thus, the gradient is \(-2\).

For the curve \(xy = 12\), differentiate implicitly: \(x \frac{dy}{dx} + y = 0\).

At \(P(2, 6)\), substitute \(x = 2\) and \(y = 6\) into the differentiated equation:

\(2 \frac{dy}{dx} + 6 = 0\)

\(\frac{dy}{dx} = -\frac{6}{2} = -3\)

The gradient of the tangent at \(P\) is \(-3\).

The angle \(\theta\) between the line and the tangent is given by:

\(\tan \theta = \left| \frac{-3 - (-2)}{1 + (-3)(-2)} \right| = \left| \frac{-1}{1 + 6} \right| = \frac{1}{7}\)

\(\theta = \tan^{-1} \left( \frac{1}{7} \right) \approx 8.1^\circ\)

(i) To find BD, use the formula for the perpendicular from the center to the chord: BD = 9\sin(\pi - 2.4). Calculating gives:

\(BD = 9\sin(\pi - 2.4) = 6.08 \text{ cm}\)

(ii) To find the perimeter of the shaded region, calculate the arc length AB and add the lengths of BD, OD, and OA:

Arc AB = 9 \times 2.4 = 21.6 \text{ cm}

Using Pythagoras or trigonometry, OD = 9\cos(\pi - 2.4) = 6.64 \text{ cm}

Perimeter = 21.6 + 6.08 + 9 + 6.64 = 43.3 \text{ cm}

(iii) To find the area of the shaded region, calculate the area of the sector and subtract the area of triangle OBD:

Area of sector = \(\frac{1}{2} \times 9^2 \times 2.4\)

Area of triangle OBD = \(\frac{1}{2} \times 6.08 \times 6.64\)

Total area = \(\frac{1}{2} \times 81 \times 2.4 - \frac{1}{2} \times 6.08 \times 6.64 = 117 \text{ cm}^2\)