To find the coordinates of \(Q\), substitute \(y = x + 4\) into the curve equation:

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

Simplify to:

\(2x^2 - 5x - 3 = 0\)

Factorize:

\((2x + 1)(x - 3) = 0\)

Solutions are \(x = -\frac{1}{2}\) and \(x = 3\). Since \(P = (3, 7)\), \(Q = \left( -\frac{1}{2}, \frac{3}{2} \right)\).

For part (iii), find the mid-point of \(AP\):

\(\left( \frac{3 + (-\frac{1}{2})}{2}, \frac{7 + \frac{3}{2}}{2} \right) = (2, 3)\)

Gradient of line \(Q\) to mid-point:

\(\frac{3/2 - 3}{-1/2 - 2} = \frac{-3/2}{-5/2} = \frac{1}{5}\)

Equation of line:

\(y - 3 = -\frac{1}{5}(x - 2)\)