(a) Rewrite the circle equation in standard form: \((x-4)^2 + (y+2)^2 = 16 + 4 + 5\). The centre is \((4, -2)\) and the radius is \(\sqrt{25} = 5\).

(b) The gradient from P (1, 2) to C (4, -2) is \(-\frac{4}{3}\). The tangent at P has a gradient of \(\frac{3}{4}\). The equation is \(y - 2 = \frac{3}{4}(x - 1)\) or \(4y = 3x + 5\).

(c) Q has the same y-coordinate as P, so \(y = 2\). Q is as far to the right of C as P is to the left, so \(x = 7\). Thus, Q is (7, 2).

(d) The gradient of the tangent at Q is \(-\frac{3}{4}\). The equation of the tangent at Q is \(y - 2 = -\frac{3}{4}(x - 7)\) or \(4y + 3x = 29\). Solving \(4y = 3x + 5\) and \(4y + 3x = 29\) gives \(T\) as \(\left(4, \frac{17}{4}\right)\).