Answer: \(CAB=0.927\) radians; perimeter \(33.3\) cm; area \(71.9\text{ cm}^2\).

(a) In triangle \(ABC\),

\(AC=3,\qquad BC=4,\qquad AB=5.\)

So triangle \(ABC\) is right-angled at \(C\). For angle \(CAB\),

\(\tan(CAB)=\frac{BC}{AC}=\frac43.\)

Therefore

\(CAB=\tan^{-1}\left(\frac43\right)=0.927\)

radians, to 3 significant figures.

(b) Let

\(\theta=CAB=0.927.\)

The angle for the outer arc on the circle with radius \(3\) is

\(2\pi-2\theta.\)

Since the angle at \(B\) in triangle \(ABC\) is \(\frac{\pi}{2}-\theta\), the corresponding angle for the smaller right-circle arc inside the overlap is

\(2\left(\frac{\pi}{2}-\theta\right)=1.287.\)

Therefore the outer arc on the circle with radius \(4\) has angle

\(2\pi-1.287.\)

Using \(s=r\theta\), the perimeter is

\(3(2\pi-2(0.927))+4(2\pi-1.287).\)

So

\(\text{perimeter}=13.287+19.985=33.3\)

cm, to 3 significant figures.

(c) The whole shape is made from two right-angled triangles and two sectors.

The area of the two right-angled triangles is

\(2\left(\frac12\cdot3\cdot4\right)=12.\)

The sector areas are

\(\frac{3^2}{2}(2\pi-2(0.927)) +\frac{4^2}{2}(2\pi-1.287).\)

Thus

\(\text{sector area}=19.93+39.97.\)

Therefore the total area is

\(12+19.93+39.97=71.9\)

\(\text{cm}^2\), to 3 significant figures.