Step-by-step Solution:

(i) Find the equation of line AB:

The gradient of the perpendicular bisector is \(\frac{3}{2}\). Therefore, the gradient of line AB is the negative reciprocal, \(-\frac{2}{3}\).

Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the gradient and \((x_1, y_1)\) is a point on the line:

\(y - 6 = -\frac{2}{3}(x + 2)\)

Simplifying:

\(y - 6 = -\frac{2}{3}x - \frac{4}{3}\)

\(y = -\frac{2}{3}x + \frac{14}{3}\)

Multiplying through by 3 to eliminate fractions:

\(3y = -2x + 14\)

Rearranging gives:

\(3y + 2x = 14\)

(ii) Find the coordinates of point B:

The midpoint of AB is the point where the perpendicular bisector intersects AB. Solving the equations \(2y = 3x + 5\) and \(3y + 2x = 14\) simultaneously:

From \(2y = 3x + 5\), express \(y\):

\(y = \frac{3}{2}x + \frac{5}{2}\)

Substitute into \(3y + 2x = 14\):

\(3\left(\frac{3}{2}x + \frac{5}{2}\right) + 2x = 14\)

\(\frac{9}{2}x + \frac{15}{2} + 2x = 14\)

\(\frac{13}{2}x + \frac{15}{2} = 14\)

\(\frac{13}{2}x = \frac{13}{2}\)

\(x = 1\)

Substitute \(x = 1\) back into \(y = \frac{3}{2}x + \frac{5}{2}\):

\(y = \frac{3}{2}(1) + \frac{5}{2} = 4\)

The midpoint is \((1, 4)\). Since A is \((-2, 6)\), use the midpoint formula to find B:

\(\left(\frac{-2 + x}{2}, \frac{6 + y}{2}\right) = (1, 4)\)

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

\(\frac{6 + y}{2} = 4 \Rightarrow y = 2\)

Thus, the coordinates of B are \((4, 2)\).