Answer: (a) \(\theta=0.8\); (b) \(3.28\text{ cm}^2\); (c) \(21.8\text{ cm}\).

(a) The arc length is given by

\(s=r\theta.\)

Here \(s=9.6\) and \(r=12\), so

\(9.6=12\theta.\)

Hence

\(\theta=0.8.\)

(b) Since \(OM\) is perpendicular to \(AB\), the angle \(AOM\) is

\(\frac{\theta}{2}=0.4.\)

In right-angled triangle \(OAM\),

\(\tan0.4=\frac{AM}{12}.\)

So

\(AM=12\tan0.4=5.074\ldots.\)

The area of triangle \(OAB\) is

\(\frac12(2AM)(12)=12AM.\)

Thus

\(\text{area of triangle }OAB=12(5.074\ldots)=60.88\ldots.\)

The area of sector \(OCD\) is

\(\frac12r^2\theta=\frac12(12^2)(0.8)=57.6.\)

The shaded area is the triangle area minus the sector area:

\(60.88\ldots-57.6=3.28\ldots.\)

Therefore the total shaded area is

\(3.28\text{ cm}^2\)

to 3 significant figures.

(c) From the same right-angled triangle,

\(\sin0.4=\frac{AM}{OA}.\)

So

\(OA=\frac{AM}{\sin0.4}=13.03\ldots.\)

Since \(OC=12\),

\(AC=OA-OC=13.03\ldots-12=1.03\ldots.\)

Similarly, \(BD=1.03\ldots\).

The total perimeter of the two shaded regions is made from \(AC\), \(BD\), the arc \(CD\), and the base \(AB=2AM\):

\(P=2(1.03\ldots)+9.6+2(5.074\ldots).\)

Therefore

\(P=21.8\text{ cm}\)

to 3 significant figures.