(a) The rotation matrix for 60° anticlockwise is \(\begin{pmatrix} \cos 60^{\circ} & -\sin 60^{\circ} \\ \sin 60^{\circ} & \cos 60^{\circ} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

The stretch matrix is \(\begin{pmatrix} d & 0 \\ 0 & 1 \end{pmatrix}\).

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

(b) The area of the parallelogram is given by the determinant of \(\textbf{M}\), which is \(\frac{1}{2}d^2\). Therefore, \(\frac{1}{2}d^2 = \frac{1}{2} \Rightarrow d^2 = 1 \Rightarrow d = 2\) since \(d \neq 0\).

(c) To find \(\textbf{N}\), we use \(\textbf{MN} = \begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}\).

First, find \(\textbf{M}^{-1}\):

\(\textbf{M}^{-1} = \frac{1}{2} \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & 1 \end{pmatrix}\).

Then, \(\textbf{N} = \textbf{M}^{-1} \begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 1 + \sqrt{3} & 1 + \sqrt{3} \\ 1 - \sqrt{3} & 1 - \sqrt{3} \end{pmatrix}\).

(d) The invariant lines satisfy \(\begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}\).

This gives \(\frac{1}{2}x + \frac{1}{2}y = mx\) and \(\frac{1}{2}x + \frac{1}{2}y = my\).

Solving, we find \(y = \frac{1}{2}x\) and \(y = -x\).