(a) Using the relationship \(\sin \theta = \frac{r}{OC} \\), we have \(OC = \frac{r}{\sin \theta} \\). Therefore, \(CD = r + \frac{r}{\sin \theta} \\).

(b) The radius of arc AB is \(4 + \frac{4}{\sin \frac{\pi}{6}} = 4 + 8 = 12 \\). The arc length \(AB = 12 \times \frac{2\pi}{6} = 4\pi \\). Thus, the perimeter is \(24 + 4\pi \\).

(c) The area of \(\triangle FOC = \frac{1}{2} \times 4 \times OC \times \sin \frac{\pi}{3} = 8\sqrt{3} \\). The area of sector \(FOE = \frac{1}{2} \times \frac{2\pi}{3} \times 4^2 = \frac{16\pi}{3} \\). Therefore, the shaded area is \(16\sqrt{3} - \frac{16\pi}{3} \\).