Answer: (a) \(30\pi\text{ cm}^2\). (b) \(\frac{7\pi}{4}+14\sin\frac{\pi}{8}\approx 10.9\text{ cm}\).

(a) The perimeter of a sector is

\(2r+r\theta.\)

Here \(r=6\), so

\(12+6\theta=2(6+5\pi).\)

Thus

\(12+6\theta=12+10\pi,\)

and hence

\(\theta=\frac{5\pi}{3}.\)

The area of the sector is

\(\frac12r^2\theta=\frac12(6^2)\left(\frac{5\pi}{3}\right)=30\pi.\)

So the area is

\(\boxed{30\pi\text{ cm}^2}.\)

(b) The shaded perimeter consists of the arc \(AB\) and the chord \(AB\).

The arc length is

\(r\theta=7\cdot\frac{\pi}{4}=\frac{7\pi}{4}.\)

The chord length is

\(2r\sin\frac{\theta}{2}=14\sin\frac{\pi}{8}.\)

Therefore the perimeter is

\(\frac{7\pi}{4}+14\sin\frac{\pi}{8}.\)

Numerically, this is approximately

\(\boxed{10.9\text{ cm}}.\)