Given the equations:

1. \(y^2 + 2x = 13\)

2. \(2y + x = 8\)

From equation (2), express \(x\) in terms of \(y\):

\(x = 8 - 2y\)

Substitute \(x = 8 - 2y\) into equation (1):

\(y^2 + 2(8 - 2y) = 13\)

Simplify:

\(y^2 + 16 - 4y = 13\)

\(y^2 - 4y + 3 = 0\)

Factor the quadratic equation:

\((y - 3)(y - 1) = 0\)

Thus, \(y = 3\) or \(y = 1\).

For \(y = 3\):

\(x = 8 - 2(3) = 2\)

Point: \((2, 3)\)

For \(y = 1\):

\(x = 8 - 2(1) = 6\)

Point: \((6, 1)\)

Therefore, the points of intersection are \((2, 3)\) and \((6, 1)\).

Given the equations:

Line: \(2y = x + 5\)

Curve: \(y = x^2 - 4x + 7\)

Substitute \(y\) from the line equation into the curve equation:

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

Substitute into the curve equation:

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

Multiply through by 2 to eliminate the fraction:

\(x + 5 = 2x^2 - 8x + 14\)

Rearrange to form a quadratic equation:

\(2x^2 - 9x + 9 = 0\)

Factor or use the quadratic formula to solve for \(x\):

\(x = 3\) or \(x = \frac{1}{2}\)

(i) To find the points of intersection, substitute \(y = 3 - x\) from the line equation into the curve equation:

\(2x^2 - 8x + 9 = 3 - x\)

Rearrange to form a quadratic equation:

\(2x^2 - 7x + 6 = 0\)

Factorize:

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

Thus, \(x = 2\) or \(x = \frac{3}{2}\).

(ii) To find the stationary point of C, differentiate \(y = 2x^2 - 8x + 9\):

\(\frac{dy}{dx} = 4x - 8\)

Set \(\frac{dy}{dx} = 0\):

\(4x - 8 = 0\)

\(x = 2\)

Therefore, \(x = 2\) is a stationary point of C.

Given the equations:

1. Curve: \(xy = 12\)

2. Line: \(2x + y = 11\)

Substitute \(y = 11 - 2x\) from the line equation into the curve equation:

\(x(11 - 2x) = 12\)

\(11x - 2x^2 = 12\)

Rearrange to form a quadratic equation:

\(2x^2 - 11x + 12 = 0\)

Factor the quadratic equation:

\((2x - 3)(x - 4) = 0\)

Thus, \(x = \frac{3}{2}\) or \(x = 4\).

For \(x = \frac{3}{2}\), substitute back to find \(y\):

\(y = 11 - 2 \times \frac{3}{2} = 8\)

For \(x = 4\), substitute back to find \(y\):

\(y = 11 - 2 \times 4 = 3\)

Therefore, the points of intersection are \((1.5, 8)\) and \((4, 3)\).

To find the points of intersection, substitute \(y = 11 - 2x\) from the line equation into the curve equation \(xy = 12\).

Substitute: \(x(11 - 2x) = 12\).

Expand and rearrange: \(11x - 2x^2 = 12\).

Rearrange to form a quadratic equation: \(2x^2 - 11x + 12 = 0\).

Factor the quadratic: \((2x - 3)(x - 4) = 0\).

Solve for \(x\): \(x = \frac{3}{2}\) or \(x = 4\).

Substitute back to find \(y\):

For \(x = \frac{3}{2}\), \(y = 11 - 2 \times \frac{3}{2} = 8\).

For \(x = 4\), \(y = 11 - 2 \times 4 = 3\).

Thus, the points of intersection are \(\left( \frac{1}{2}, 8 \right)\) and \((4, 3)\).