(a) The equation of the circle is \((x - 7)^2 + (y + 4)^2 = 29\). The centre \(C\) is \((7, -4)\). The line \(y = -2\) intersects the circle, substituting \(y = -2\) into the circle's equation gives:

\((x - 7)^2 + (-2 + 4)^2 = 29\)

\((x - 7)^2 + 4 = 29\)

\((x - 7)^2 = 25\)

\(x - 7 = \pm 5\)

\(x = 12\) or \(x = 2\)

Thus, \(A = (2, -2)\) and \(B = (12, -2)\).

(b) Using the cosine rule in triangle \(ACB\):

\(\cos \theta = \frac{29 + 29 - 100}{2 \times \sqrt{29} \times \sqrt{29}} = \frac{-21}{29}\)

\(\theta = \cos^{-1}\left(\frac{-21}{29}\right)\)

\(\theta = 2.38\) radians (to 3 significant figures).

(c) The length of \(AC = BC = \sqrt{29}\). The area of the larger segment is:

Area of circle sector = \(\frac{1}{2} \times 29 \times (2\pi - 2.38) = 56.587\)

Area of triangle \(ACB = \frac{1}{2} \times \sqrt{29} \times \sqrt{29} \times \sin(2.38) = 10\)

Area of larger segment = \(\pi \times 29 - 34.52 + 10 = 66.6\)