(a) The stretch parallel to the x-axis with scale factor 14 is represented by the matrix \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}\).

The rotation anticlockwise about the origin through angle \(\frac{1}{3} \pi\) is represented by \(\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

Thus, \(\mathbf{M} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & -\frac{\sqrt{3}}{2} \\ 7\sqrt{3} & \frac{1}{2} \end{pmatrix}\).

Therefore, \(2\mathbf{M} = \begin{pmatrix} 14 & -\sqrt{3} \\ 14\sqrt{3} & 1 \end{pmatrix}\).

(b) The transformation \(\mathbf{M}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} 7x - \frac{\sqrt{3}}{2}y \\ 7\sqrt{3}x + \frac{1}{2}y \end{pmatrix}\).

For invariant lines, \(y = mx\) and \(Y = mX\).

\(7\sqrt{3}x + \frac{1}{2}mx = m(7x - \frac{\sqrt{3}}{2}mx)\).

\(7\sqrt{3} + \frac{1}{2}m = 7m - \frac{\sqrt{3}}{2}m^2\).

\(\sqrt{3}m^2 - 13m + 14\sqrt{3} = 0\).

Solving gives \(m = 2\sqrt{3}\) and \(m = \frac{7}{\sqrt{3}}\).

Thus, the invariant lines are \(y = 2\sqrt{3}x\) and \(y = \frac{1}{\sqrt{3}}x\).

(c) The inverse of \(2\mathbf{M}\) is \(\mathbf{M}^{-1} = \frac{1}{14} \begin{pmatrix} 1 & \sqrt{3} \\ -14\sqrt{3} & 7 \end{pmatrix}\).