(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:

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