(i) The gradient of AB is calculated as:

\(\frac{22 - (-2)}{15 - 3} = \frac{24}{12} = 2\)

Since the gradient of AB is given as 2m, equating gives:

\(2m = 2\)

Thus, \(m = 1\).

(ii) The equation of line AC is:

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

\(y = -2x + 6 - 2\)

\(y = -2x + 4\)

The equation of line BC is:

\(y - 22 = 1(x - 15)\)

\(y = x + 7\)

Solving the simultaneous equations:

\(y = -2x + 4\)

\(y = x + 7\)

Equating gives:

\(-2x + 4 = x + 7\)

\(-3x = 3\)

\(x = -1\)

Substitute \(x = -1\) into \(y = x + 7\):

\(y = -1 + 7 = 6\)

Thus, \(C = (-1, 6)\).

(iii) The midpoint M of AB is:

\(M = \left( \frac{3 + 15}{2}, \frac{-2 + 22}{2} \right) = (9, 10)\)

The gradient of the perpendicular bisector is:

\(-\frac{1}{2}\)

The equation of the perpendicular bisector is:

\(y - 10 = -\frac{1}{2}(x - 9)\)

\(2y - 20 = -x + 9\)

\(2y + x = 29\)

Solving the equations \(y = x + 7\) and \(2y + x = 29\):

Substitute \(y = x + 7\) into \(2y + x = 29\):

\(2(x + 7) + x = 29\)

\(2x + 14 + x = 29\)

\(3x = 15\)

\(x = 5\)

Substitute \(x = 5\) into \(y = x + 7\):

\(y = 5 + 7 = 12\)

Thus, \(D = (5, 12)\).