(a) \(DB=3\), so \(DC=\sqrt{3^2+4^2}=5\), equal to \(DA\). Let \(\phi=\angle ADC=\pi-\tan^{-1}(4/3)=2.2142974\ldots\). Arc \(AC=5\phi\), and chord \(AC=\sqrt{8^2+4^2}=\sqrt{80}\). The segment perimeter is \(\boxed{5\phi+\sqrt{80}\approx20.0\text{ cm}}\) (3 s.f.).

(b) The sector area is \(\tfrac12(5^2)\phi\); triangle \(ADC\) has base \(AD=5\) and perpendicular height \(4\), so area \(10\). Thus \(\boxed{A=12.5\phi-10\approx17.7\text{ cm}^2}\) (3 s.f.).