(a) To find angle BAO, use the cosine rule in triangle AOB:

\(\cos BAO = \frac{8^2 + 12^2 - 8^2}{2 \times 8 \times 12}\)

\(\cos BAO = \frac{6}{8}\)

Thus, \(BAO = 0.723\) radians.

(b) The area of the shaded region is the area of sector ABC minus the area of triangle AOB.

Area of sector ABC:

\(\frac{1}{2} \times 12^2 \times 0.7227 = 52.0\)

Area of triangle AOB:

\(\frac{1}{2} \times 8 \times 12 \times \sin(0.7227) = 31.7\)

Shaded area:

\(52.0 - 31.7 = 20.3\)

(c) The perimeter of the shaded region is the sum of arc BC and line segments OC and AB.

Arc BC:

\(12 \times 0.7227 = 8.67\)

Perimeter:

\(8 + 4 + 8.67 = 20.7\)