(i) The midpoint M of AC is calculated as:

\(M = \left( \frac{-3 + 5}{2}, \frac{2 + 6}{2} \right) = (1, 4)\)

The gradient of AC is:

\(\text{Gradient of } AC = \frac{6 - 2}{5 + 3} = \frac{1}{2}\)

The perpendicular bisector has a gradient of:

\(-2\)

The equation of MB is:

\(y - 4 = -2(x - 1)\)

Simplifying gives:

\(y = -2x + 6\)

Setting \(y = 0\) to find B:

\(0 = -2x + 6\) \(x = 3\)

So, B is (3, 0).

(ii) The gradient of AB is:

\(\frac{2 - 0}{-3 - 3} = -\frac{2}{6} = -\frac{1}{3}\)

The gradient of BC is:

\(\frac{6 - 0}{5 - 3} = \frac{6}{2} = 3\)

Since:

\(-\frac{1}{3} \times 3 = -1\)

AB is perpendicular to BC.

(iii) Since ABCD is a square, D is calculated as:

\(D = (-1, 8)\)

The length of AD is:

\(AD = \sqrt{(-1 + 3)^2 + (8 - 2)^2} = \sqrt{4 + 36} = \sqrt{40}\)

or approximately 6.32.