(a) Since the total angle around point O is 2π radians, and angle AOB is 2.8 radians, angle ACO is given by:

\(\angle ACO = \pi - 2.8 = 0.7 \text{ radians}\)

(b) Using the sine rule in triangle ACO, we have:

\(\frac{R}{\sin(2.8)} = \frac{r}{\sin(0.7)}\)

Solving for R gives:

\(R = \frac{r \cdot \sin(2.8)}{\sin(0.7)} \approx 1.53r\)

(c) The area of sector OAB is:

\(\text{Area of sector } OAB = \frac{1}{2} r^2 \times 2.8 = 1.4r^2\)

The area of sector CAB is:

\(\text{Area of sector } CAB = \frac{1}{2} R^2 \times 2 \times 0.7 = 1.638r^2\)

The area of triangle OAC is:

\(\text{Area of } \triangle OAC = \frac{1}{2} r^2 \sin(\pi - 2.8) = 0.4927r^2\)

The area of the shaded region is:

\(1.4r^2 - (1.638r^2 - 0.985r^2) = 0.747r^2 \text{ to } 0.748r^2\)