Answer: (a) \(r=3\). (b) \(6\pi-\frac{9\sqrt3}{2}\text{ cm}^2\).

The distance \(C_1C_2\) is a radius of both circles, so

\(C_1C_2=C_1A=C_2A=r.\)

Thus triangle \(C_1AC_2\) is equilateral, and the angle at each centre between the two intersection points is

\(\frac{2\pi}{3}.\)

(a) The perimeter of the shaded lens consists of two equal arcs, each subtending angle \(\frac{2\pi}{3}\) at a centre. Hence

\(\text{perimeter}=2\left(r\cdot\frac{2\pi}{3}\right)=\frac{4\pi r}{3}.\)

Given this perimeter is \(4\pi\),

\(\frac{4\pi r}{3}=4\pi.\)

Therefore

\(\boxed{r=3}.\)

(b) The shaded area is twice the area of a sector of angle \(\frac{2\pi}{3}\), minus twice the area of an equilateral triangle of side \(r\).

One sector has area

\(\frac12r^2\cdot\frac{2\pi}{3}=\frac{\pi r^2}{3}.\)

One equilateral triangle has area

\(\frac{\sqrt3}{4}r^2.\)

Therefore the shaded area is

\(2\left(\frac{\pi r^2}{3}-\frac{\sqrt3}{4}r^2\right).\)

Using \(r=3\),

\(2\left(3\pi-\frac{9\sqrt3}{4}\right)=6\pi-\frac{9\sqrt3}{2}.\)

So the exact shaded area is

\(\boxed{6\pi-\frac{9\sqrt3}{2}\text{ cm}^2}.\)