(a) In triangle \(AOB\), \(AO=OB=8\) and \(AB=12\). By the cosine rule, \[\cos\alpha=\frac{8^2+12^2-8^2}{2(8)(12)}=\frac34,\qquad\alpha=\angle BAO=\cos^{-1}(3/4)\approx\boxed{0.723\text{ rad}}.\]

(b) Subtract triangle \(AOB\) from sector \(ABC\): \[A=\frac12(12^2)\alpha-\frac12(8)(12)\sin\alpha=72\cos^{-1}(3/4)-12\sqrt7\approx\boxed{20.3\text{ cm}^2}.\]

(c) \(OC=AC-AO=12-8=4\). The boundary is \(OB+OC+\text{arc }BC\), so \[P=8+4+12\alpha\approx\boxed{20.7\text{ cm}}.\]