(a) To find the coordinates of \(P\), equate the line and curve equations: \(mx + c = -\frac{m}{x}\). Rearrange to get \(mx^2 + cx + m = 0\). For the line to be tangent, the discriminant must be zero: \(b^2 - 4ac = 0\), giving \(c^2 - 4m^2 = 0\). Solving gives \(c = \pm 2m\). Substitute \(c = 2m\) into the quadratic: \(mx^2 + 2mx + m = 0\), which simplifies to \((x+1)^2 = 0\), so \(x = -1\). Substituting back, \(y = m\), so \(P = (-1, m)\).

(b) The equation of the normal at \(P\) is \(y - m = -\frac{1}{m}(x + 1)\). Simplifying gives \(y = -\frac{1}{m}x + \frac{1}{m} + m\). Equate this to the curve equation \(y = -\frac{m}{x}\) to find \(x\): \(-\frac{1}{m}x + \frac{1}{m} + m = -\frac{m}{x}\). Rearrange to get \(x^2(1 - m^2) - m^2 = 0\). Solving gives \(x = m^2\). Substitute back to find \(y\): \(y = -\frac{1}{m}\), so \(Q = (m^2, -\frac{1}{m})\).