(a) To find the shortest distance between \(l_1\) and \(l_2\), we first find the direction vectors of the lines: \(\mathbf{d_1} = \begin{pmatrix} -4 \\ 3 \\ 5 \end{pmatrix}\) and \(\mathbf{d_2} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}\).

The vector between a point on \(l_1\) and a point on \(l_2\) is \(\mathbf{b} = \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix}\).

The cross product \(\mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 5 \\ 2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix}\).

The shortest distance is given by \(\frac{|\mathbf{b} \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|} = \frac{67}{\sqrt{139}} \approx 5.68\).

(b) The plane \(\Pi\) contains \(l_1\) and the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\). The normal vector to the plane is \(\mathbf{n} = \mathbf{d_1} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -4 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 3 \\ -1 \\ -1 \end{pmatrix}\).

The equation of the plane is \(1(-1) + 3(-3) - 1(-4) = -6 \Rightarrow x + 3y - z = -6\).