(a) The gradient of \(AB\) is \(-1\) because the tangents have gradient 1 and are perpendicular to the radius. The center of the circle is \((4, -1)\). Using the point-slope form, the equation of \(AB\) is \(y + 1 = -1(x - 4)\), which simplifies to \(y = -x + 3\).

(b) Substitute \(y = -x + 3\) into the circle equation \((x-4)^2 + (y+1)^2 = 40\):

\((x-4)^2 + (-x+3+1)^2 = 40\)

\(2(x-4)^2 = 40\) or \((x-4)^2 = 20\)

\(x = 4 \pm \sqrt{20}\)

Coordinates of \(A\) are \((4 - \sqrt{20}, -1 + \sqrt{20})\).

(c) The equation of the tangent at \(A\) is found using the point \((4 - \sqrt{20}, -1 + \sqrt{20})\) and gradient 1:

\(y - (-1 + \sqrt{20}) = 1(x - (4 - \sqrt{20}))\)

\(y = x - 5 + 4\sqrt{5}\)