(a) The midpoint of AB is the center of the circle C. Calculate the midpoint: \(\left(\frac{-1+7}{2}, \frac{-2+4}{2}\right) = (3, 1)\).

The radius is half the distance of AB. Calculate the distance: \(\sqrt{(7 - (-1))^2 + (4 - (-2))^2} = \sqrt{64 + 36} = 10\). Thus, the radius is 5.

The equation of the circle is \((x-3)^2 + (y-1)^2 = 5^2 = 25\).

(b) The gradient of AB is \(\frac{4 - (-2)}{7 - (-1)} = \frac{6}{8} = \frac{3}{4}\). The gradient of the tangent T is the negative reciprocal: \(-\frac{4}{3}\).

The equation of the tangent at B is \(y - 4 = -\frac{4}{3}(x - 7)\). Simplifying gives \(3y + 4x = 40\).

(c) The reflection of the circle C in the line T has its center at \((11, 7)\) (since B is the midpoint of the line joining the centers).

The radius remains the same, so the new equation is \((x-11)^2 + (y-7)^2 = 25\).