(i) To find angle BAO, use the tangent function: \(\tan \theta = \frac{5}{12}\). Solving for \(\theta\), we get \(\theta = \tan^{-1}\left(\frac{5}{12}\right) \approx 0.3948\) radians.

(ii) The area of triangle AOB is \(\frac{1}{2} \times 12 \times 5 = 30\) cm².

The shaded area is the sum of two sectors minus the triangle area. The area of the sector with center A is \(\frac{1}{2} \times 12^2 \times 0.3948\).

The other angle in the triangle is \(\frac{\pi}{2} - 0.3948\). The area of the sector with center B is \(\frac{1}{2} \times 5^2 \times \left(\frac{\pi}{2} - 0.3948\right)\).

Thus, the shaded area is:

\(\frac{1}{2} \times 12^2 \times 0.3948 + \frac{1}{2} \times 5^2 \times \left(\frac{\pi}{2} - 0.3948\right) - 30\)

\(= 28.43 + 14.70 - 30 = 13.1\) cm².