1. The midpoint of the diagonals is given as \(E \left( 6 \frac{1}{2}, 8 \frac{1}{2} \right)\).

2. Use the midpoint formula for diagonals: \(\frac{x_1 + x_3}{2} = 6.5\) and \(\frac{y_1 + y_3}{2} = 8.5\).

3. Solve the equations for \(A\) and \(C\):

\(y = 3x\) and \(4y = x + 11\).

4. Solving these simultaneously gives \(A(1, 3)\).

5. Use the midpoint formula to find \(C\):

\(x_3 = 2 \times 6.5 - 1 = 12\) and \(y_3 = 2 \times 8.5 - 3 = 14\).

6. Thus, \(C(12, 14)\).

7. Find the equation of \(BC\) using the midpoint and slope:

\(4y = x + 44\).

8. Solve for \(B\) using \(y = 3x\) and \(4y = x + 44\):

\(B(4, 12)\).

9. Use the midpoint formula to find \(D\):

\(x_4 = 2 \times 6.5 - 4 = 9\) and \(y_4 = 2 \times 8.5 - 12 = 5\).

10. Thus, \(D(9, 5)\).

Given the points A (0, 8) and B (4, 0), and the equation of diagonal AC as \(8y + x = 64\).

First, find the slope of AB: \(m_{AB} = \frac{8 - 0}{0 - 4} = -2\).

Since ABCD is a rectangle, BC is perpendicular to AB, so \(m_{BC} \times m_{AB} = -1\).

Thus, \(m_{BC} = \frac{1}{2}\).

The equation of line BC is \(y - 0 = \frac{1}{2}(x - 4)\), which simplifies to \(y = \frac{1}{2}x - 2\).

Now, solve the equations \(8y + x = 64\) and \(y = \frac{1}{2}x - 2\) simultaneously:

Substitute \(y = \frac{1}{2}x - 2\) into \(8y + x = 64\):

\(8(\frac{1}{2}x - 2) + x = 64\)

\(4x - 16 + x = 64\)

\(5x = 80\)

\(x = 16\)

Substitute \(x = 16\) back into \(y = \frac{1}{2}x - 2\):

\(y = \frac{1}{2}(16) - 2 = 8 - 2 = 6\)

Thus, the coordinates of C are (16, 6).

For D, use the vector step method or the fact that AD is parallel to BC:

Since AD is parallel to BC, \(m_{AD} = \frac{1}{2}\).

Using the midpoint M of AC as (8, 7), and knowing D is symmetric to C about M, calculate:

\(D = (2 \times 8 - 16, 2 \times 7 - 6) = (12, 14)\).

Thus, the coordinates of D are (12, 14).

(i) To find the coordinates of \(X\), we first calculate the slope of \(AB\):

\(m_{AB} = \frac{6 - 14}{14 - 2} = -\frac{2}{3}\).

The slope of \(CX\), being perpendicular to \(AB\), is the negative reciprocal:

\(m_{CX} = \frac{3}{2}\).

The equation of line \(AB\) is:

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

The equation of line \(CX\) is:

\(y - 2 = \frac{3}{2}(x - 7)\).

Solving these equations simultaneously gives the coordinates of \(X\):

\(X (11, 8)\).

(ii) To find the ratio \(AX : XB\), calculate the distances:

\(AX = \sqrt{(11 - 2)^2 + (8 - 14)^2} = \sqrt{9^2 + 6^2} = \sqrt{117}\).

\(XB = \sqrt{(14 - 11)^2 + (6 - 8)^2} = \sqrt{3^2 + 2^2} = \sqrt{13}\).

The ratio \(AX : XB = \sqrt{117} : \sqrt{13} = 3:1\).

(i) To find the equation of BC, we first calculate the gradient of AB. The gradient \(m\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For AB, \(m = \frac{11 - 3}{5 - 1} = 2\).

Since ABC is a right angle, the gradient of BC is the negative reciprocal of the gradient of AB, so \(m_{BC} = -\frac{1}{2}\).

The equation of a line is \(y - y_1 = m(x - x_1)\). Using point B (5, 11), the equation of BC is \(y - 11 = -\frac{1}{2}(x - 5)\).

(ii) To find the coordinates of C, we first find the equation of AC. The gradient of AC (or AX) is \(\frac{1}{3}\) since X (4, 4) lies on AC.

The equation of AC using point A (1, 3) is \(y - 3 = \frac{1}{3}(x - 1)\).

Alternatively, using point X (4, 4), the equation is \(y - 4 = \frac{1}{3}(x - 4)\).

Solving the simultaneous equations of BC and AC gives the coordinates of C as (13, 7).

(i) To find the coordinates of M, we first find the equations of lines AC and BD. The gradient of AC is \(\frac{4 - (-1)}{9 - (-1)} = \frac{1}{2}\). The equation of line AC is \(y + 1 = \frac{1}{2}(x + 1)\).

The gradient of BD is \(-2\) because it is perpendicular to AC (\(m_1 m_2 = -1\)). The equation of line BD is \(y - 6 = -2(x - 3)\).

Solving these equations simultaneously gives the intersection point M as (5, 2).

\(To find D, use the midpoint formula or vector method. Since BM = MD, D is at (7, -2).\)

(ii) To find the ratio AM : MC, calculate the distances using the distance formula. \(AM = \sqrt{(5 - (-1))^2 + (2 - (-1))^2} = \sqrt{45}\) and \(MC = \sqrt{(9 - 5)^2 + (4 - 2)^2} = \sqrt{20}\).

The ratio is \(\sqrt{45} : \sqrt{20} = 3 : 2\).