Answer: (a) \(\frac{\pi}{3}\); (b) \(4a+\frac{\pi a}{3}\); (c) \(\sqrt3a^2-\frac{\pi a^2}{6}\).

(a) The radius of the large circle is \(2a\), and the chord \(AB\) has length \(2a\).

For a chord of length \(2a\),

\(2(2a)\sin\frac{\angle AOB}{2}=2a.\)

So

\(\sin\frac{\angle AOB}{2}=\frac12.\)

Hence

\(\frac{\angle AOB}{2}=\frac{\pi}{6},\)

and therefore

\(\angle AOB=\frac{\pi}{3}.\)

(b) Since \(C\) and \(D\) are mid-points of \(OB\) and \(OA\),

\(AD=a,\qquad BC=a.\)

The length of the arc \(DC\), with centre \(O\), radius \(a\), and angle \(\frac{\pi}{3}\), is

\(a\cdot\frac{\pi}{3}=\frac{\pi a}{3}.\)

The perimeter of the shaded region is therefore

\(AB+AD+BC+\text{arc }DC =2a+a+a+\frac{\pi a}{3}.\)

So the perimeter is

\(4a+\frac{\pi a}{3}.\)

(c) The shaded area is the area of triangle \(AOB\) minus the area of sector \(DOC\).

The area of triangle \(AOB\) is

\(\frac12(2a)(2a)\sin\frac{\pi}{3} =\sqrt3a^2.\)

The area of sector \(DOC\) is

\(\frac12a^2\cdot\frac{\pi}{3} =\frac{\pi a^2}{6}.\)

Therefore the shaded area is

\(\sqrt3a^2-\frac{\pi a^2}{6}.\)