(a) To find when the line is tangent to the curve, substitute \(y = mx + c\) into \(xy = 16\):

\(x(mx + c) = 16\)

\(mx^2 + cx - 16 = 0\)

For the line to be tangent, the discriminant of this quadratic must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = m\), \(b = c\), \(c = -16\).

\(c^2 - 4(m)(-16) = 0\)

\(c^2 + 64m = 0\)

\(m = -\frac{c^2}{64}\)

(b) Substitute \(m = -4\) into the equation:

\(x(-4x + c) = 16\)

\(-4x^2 + cx - 16 = 0\)

For two distinct points, the discriminant must be positive:

\(b^2 - 4ac > 0\)

\(c^2 - 4(-4)(-16) > 0\)

\(c^2 - 256 > 0\)

\(c^2 > 256\)

\(c > 16\) or \(c < -16\)