(a) The center of the circle is found by completing the square: \((x-3)^2 + (y+2)^2 = 40\), so the center is \((3, -2)\) and the radius is \(\sqrt{40}\).

The gradient of the radius at \(P (5, 4)\) is \(\frac{-2 - 4}{3 - 5} = 3\). The gradient of the tangent is the negative reciprocal, \(-\frac{1}{3}\).

The equation of the tangent line is \(y - 4 = -\frac{1}{3}(x - 5)\), which simplifies to \(y = -\frac{1}{3}x + \frac{17}{3}\).

Intercepts: \(A (17, 0)\) and \(B (0, \frac{17}{3})\).

Area of \(\triangle OAB = \frac{1}{2} \times 17 \times \frac{17}{3} = \frac{289}{6}\).

(b) The radius of the circle is \(\sqrt{40}\).

For \(\triangle PQR\) to be equilateral, each side is \(\sqrt{40}\). The area of \(\triangle PQR\) is \(\frac{\sqrt{3}}{4} \times (\sqrt{40})^2 = 30\sqrt{3}\).