(i) The angle BAC in radians is given by the formula for arc length: \(\theta = \frac{\text{arc length}}{\text{radius}}\). Here, \(\theta = \frac{4}{5} = 0.8\) radians.

(ii) To find the area of the shaded region BDC, we calculate:

1. The length of BD: \(BD = 5 \sin(0.8) \approx 3.587\) cm.

2. The length of DC: \(DC = 5 - 5 \cos(0.8) \approx 1.516\) cm.

3. The area of sector BAC: \(\frac{1}{2} \times 5^2 \times 0.8 = 10\) cm².

4. The area of segment BC: \(\frac{1}{2} \times 5^2 \times (0.8 - \sin(0.8)) \approx 1.03\) cm².

5. The area of triangle BDC: \(\frac{1}{2} \times BD \times DC \approx 2.719\) cm².

6. The area of the trapezium: \(\frac{1}{2} \times (5 + DC) \times BD \approx 11.69\) cm².

7. The shaded area BDC: \(\text{Trapezium} - \text{Sector} + \text{Segment} = 11.69 - 10 + 1.03 = 1.69\) cm².

(i) To find angle CBY, use the cosine rule in triangle ACB:

\(\cos(\angle ABC) = \frac{AC^2 + BC^2 - AB^2}{2 \times AC \times BC} = \frac{12^2 + 8^2 - 8^2}{2 \times 12 \times 8} = \frac{80}{192} = \frac{5}{12}\)

\(\angle ABC = \cos^{-1}\left(\frac{5}{12}\right) \approx 1.696 \text{ radians}\)

Since CBY is a straight line, \(\angle CBY = \pi - 2 \times \angle ABC\)

\(\angle CBY = \pi - 2 \times 1.696 \approx 1.445 \text{ radians}\)

(ii) The perimeter of the shaded region consists of arcs CY and XC:

Arc CY is part of a circle with radius 8 ext{ cm} and angle CBY:

\(\text{Arc } CY = 8 \times 1.445 = 11.56 \text{ cm}\)

To find arc XC, use angle BAC:

\(\angle BAC = \frac{\pi - \angle ABC}{2} \approx 0.7227 \text{ radians}\)

Arc XC is part of a circle with radius 12 ext{ cm}:

\(\text{Arc } XC = 12 \times 0.7227 = 8.673 \text{ cm}\)

The perimeter of the shaded region is:

\(11.56 + 8.673 + 4 = 24.2 \text{ cm}\)

First, calculate angle OAB:

\(\angle OAB = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}\)

Calculate the area of sector CAB:

\(\text{Area of sector } CAB = \frac{1}{2} \times \left(\frac{3\pi}{10}\right) \times 5^2 = 11.78\)

Calculate OA using the sine rule:

\(OA = \frac{5}{\sin \frac{\pi}{5}} = 8.507\)

Calculate the area of sector COD:

\(\text{Area of sector } COD = \frac{1}{2} \times (3.507)^2 \times \frac{\pi}{5} = 3.86\)

Calculate the area of triangle OAB:

\(\Delta OAB = \frac{1}{2} \times 5 \times 8.507 \times \sin \frac{3\pi}{10} = 17.20 \text{ or } 17.21\)

Calculate the shaded area:

\(\text{Shaded area} = 17.20 \text{ or } 17.21 - 11.78 - 3.86 = 1.56 \text{ or } 1.57\)

First, find the angle ∠AOC using the arc length formula:

Arc length = radius × angle in radians.

Given arc length = 6 cm and radius = 5 cm, we have:

6 = 5 × ∠AOC

∠AOC = \frac{6}{5} \text{ radians} \approx 1.2 \text{ radians}

Next, find the length of AB using the tangent function:

AB = 5 \times \tan(1.2) \approx 12.86 \text{ cm}

Calculate the area of triangle OAB:

\text{Area of } \triangle OAB = \frac{1}{2} \times 5 \times 12.86 \approx 32.15 \text{ cm}^2

Calculate the area of sector OAC:

\text{Area of sector OAC} = \frac{1}{2} \times 5^2 \times 1.2 \approx 15 \text{ cm}^2

Finally, find the area of the shaded region:

\text{Shaded region} = 32.15 - 15 = 17.2 \text{ cm}^2

(a) To find the area of the shaded region, we first need to find the radius r and angle θ.

We have the equations:

\(\frac{1}{2} r^2 \theta = 76.8\)

\(r \theta = 9.6\)

From the second equation, \(\theta = \frac{9.6}{r}\).

Substitute \(\theta\) in the first equation:

\(\frac{1}{2} r^2 \left( \frac{9.6}{r} \right) = 76.8\)

\(\frac{1}{2} r \times 9.6 = 76.8\)

\(4.8r = 76.8\)

\(r = 16\)

Now, substitute \(r = 16\) back to find \(\theta\):

\(\theta = \frac{9.6}{16} = 0.6 \text{ radians}\)

The area of triangle \(\Delta OAB\) is:

\(\frac{1}{2} \times 16^2 \times \sin(0.6) \approx 4.53 \text{ cm}^2\)

The area of the shaded region is:

\(76.8 - 72.27 = 4.53 \text{ cm}^2\)

(b) To find the perimeter of the shaded region, we need the length of AB and the arc length.

Using the cosine rule or sine rule, we find:

\(AB = 2 \times 16 \times \sin(0.3) \approx 9.46 \text{ cm}\)

The perimeter is:

\(9.6 + 9.46 = 19.1 \text{ cm}\)