(a) To find the equation of the plane \(\Pi\), we need a normal vector to the plane. The direction vector of \(l_1\) is \(\mathbf{i} - \mathbf{j} - 4\mathbf{k}\), and the plane is parallel to \(2\mathbf{i} + 5\mathbf{j} - 4\mathbf{k}\). The normal vector is the cross product of these two vectors:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -4 \\ 2 & 5 & -4 \end{vmatrix} = \mathbf{i}(16 - 20) - \mathbf{j}(-4 - (-8)) + \mathbf{k}(5 + 2) = -4\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\)

Using the point \((1, 3, -1)\) on \(l_1\), the equation of the plane is:

\(24(1) - 4(3) + 7(-1) = d \Rightarrow 24x - 4y + 7z = 5\).

(b) To find the acute angle between \(l_2\) and \(\Pi\), we use the dot product of the direction vector of \(l_2\), \(5\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}\), and the normal vector of \(\Pi\), \(24\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\):

\((24, -4, 7) \cdot (5, -5, -2) = 24 \times 5 + (-4) \times (-5) + 7 \times (-2) = 120 + 20 - 14 = 126\).

The magnitudes are \(\sqrt{24^2 + (-4)^2 + 7^2} = \sqrt{641}\) and \(\sqrt{5^2 + (-5)^2 + (-2)^2} = \sqrt{54}\).

\(\cos \alpha = \frac{126}{\sqrt{641} \times \sqrt{54}}\).

The angle \(\alpha\) is the angle between the line and the normal, so the acute angle between \(l_2\) and \(\Pi\) is \(90^\circ - \alpha = 42.6^\circ\).