(a) To find the equation of the tangent, first calculate the gradient of line AB:

\(m_{AB} = \frac{4 - 2}{-1 - 3} = -\frac{1}{2}\).

The gradient of the tangent is the negative reciprocal: \(2\).

Using point B (3, 2), the equation of the tangent is:

\(y - 2 = 2(x - 3)\).

(b) The radius of the circle with centre B is the distance AB:

\(AB^2 = (4 - 2)^2 + (-1 - 3)^2 = 20\).

The equation of the circle is:

\((x - 3)^2 + (y - 2)^2 = 20\).

(c) Substitute the equation of the tangent into the circle's equation:

\((x - 3)^2 + (2x - 6)^2 = 20\).

Expanding and simplifying:

\(5x^2 - 30x + 25 = 20\).

\(5(x^2 - 6x + 1) = 0\).

\((x - 5)(x - 1) = 0\).

Thus, \(x = 5\) or \(x = 1\).