(i) The gradient of \(L_1\) is \(-2\) (since \(y = -2x + 8\)).

For \(L_2\) to be perpendicular to \(L_1\), its gradient is the negative reciprocal: \(\frac{1}{2}\).

Using the point-slope form \(y - y_1 = m(x - x_1)\) with point \(A(7, 4)\):

\(y - 4 = \frac{1}{2}(x - 7)\)

(ii) Solve \(2x + y = 8\) and \(y - 4 = \frac{1}{2}(x - 7)\) simultaneously:

Substitute \(y = \frac{1}{2}(x - 7) + 4\) into \(2x + y = 8\):

\(2x + \frac{1}{2}(x - 7) + 4 = 8\)

\(2x + \frac{1}{2}x - \frac{7}{2} + 4 = 8\)

\(\frac{5}{2}x - \frac{7}{2} + 4 = 8\)

\(\frac{5}{2}x = \frac{15}{2}\)

\(x = 3\)

Substitute \(x = 3\) back to find \(y\):

\(y = \frac{1}{2}(3 - 7) + 4 = 2\)

Point \(B\) is \((3, 2)\).

Length of \(AB\):

\(AB = \sqrt{(7 - 3)^2 + (4 - 2)^2} = \sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47\)