(i) The equation of a line with gradient \(m\) passing through the point \((2, 0)\) is given by the point-slope form:

\(y - 0 = m(x - 2)\)

Thus, the equation is \(y = m(x - 2)\).

(ii) To find the values of \(m\) for which the line is tangent to the curve \(y = x^2 - 4x + 5\), equate the line equation \(y = mx - 2m\) to the curve:

\(x^2 - 4x + 5 = mx - 2m\)

Rearrange to form a quadratic equation:

\(x^2 - (4 + m)x + (5 + 2m) = 0\)

For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

\((4 + m)^2 - 4(5 + 2m) = 0\)

\(m^2 - 4 = 0\)

Solving gives \(m = \pm 2\).

For \(m = 2\):

\(x^2 - 6x + 9 = 0\)

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

\(x = 3\)

Substitute \(x = 3\) into \(y = 2(x - 2)\):

\(y = 2(3 - 2) = 2\)

Point is (3, 2).

For \(m = -2\):

\(x^2 - 2x + 1 = 0\)

\((x - 1)^2 = 0\)

\(x = 1\)

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

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

Point is (1, 2).

To find the value of \(k\) for which the line is tangent to the curve, we first eliminate \(x\) from the equations. Substitute \(x = k - 2y\) from the line equation into the curve equation:

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

Simplify to get:

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

\(y^2 - 4y + 2k - 13 = 0\)

This is a quadratic equation in \(y\). For the line to be tangent to the curve, the discriminant of this quadratic must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 1\), \(b = -4\), and \(c = 2k - 13\). So,

\((-4)^2 - 4(1)(2k - 13) = 0\)

\(16 - 8k + 52 = 0\)

\(68 = 8k\)

\(k = \frac{68}{8} = 8 \frac{1}{2}\)

Thus, the value of \(k\) for which the line is tangent to the curve is \(8 \frac{1}{2}\).

To find the values of \(k\) for which the line is tangent to the curve, we set the equations equal to each other:

\(kx + 6 = x^2 + 3x + 2k\)

Rearrange to form a quadratic equation:

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

For the line to be tangent to the curve, the discriminant of this quadratic must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 1\), \(b = 3-k\), \(c = 2k-6\).

Substitute into the discriminant formula:

\((3-k)^2 - 4(1)(2k-6) = 0\)

Simplify:

\((3-k)^2 - 8k + 24 = 0\)

\(9 - 6k + k^2 - 8k + 24 = 0\)

\(k^2 - 14k + 33 = 0\)

Factorize the quadratic:

\((3-k)(11-k) = 0\)

Thus, \(k = 3\) or \(k = 11\).

To find the values of m for which the line intersects the curve at two distinct points, we equate the equations of the line and the curve:

\(mx + 4 = 3x^2 - 4x + 7\)

Rearrange to form a quadratic equation:

\(3x^2 - (4 + m)x + 3 = 0\)

For the line to intersect the curve at two distinct points, the discriminant of this quadratic equation must be greater than zero:

\(b^2 - 4ac > 0\)

Here, \(a = 3\), \(b = -(4 + m)\), and \(c = 3\).

Calculate the discriminant:

\((4 + m)^2 - 4 \times 3 \times 3 > 0\)

\((4 + m)^2 - 36 > 0\)

\((4 + m)^2 > 36\)

Taking the square root of both sides:

\(|4 + m| > 6\)

This gives two inequalities:

\(4 + m > 6\) or \(4 + m < -6\)

Solving these:

\(m > 2\) or \(m < -10\)

Thus, the set of values of m for which the line intersects the curve at two distinct points is \(m > 2 \text{ or } m < -10\).

(i) Since the roots of the equation are \(-3\) and \(5\), we can express the quadratic as \((x + 3)(x - 5) = 0\).

Expanding this, we get:

\(x^2 - 5x + 3x - 15 = x^2 - 2x - 15\).

Thus, \(p = -2\) and \(q = -15\).

(ii) For the equation \(x^2 + px + q + r = 0\) to have equal roots, the discriminant must be zero.

The discriminant \(b^2 - 4ac = 0\).

Here, \(a = 1\), \(b = p = -2\), and \(c = q + r = -15 + r\).

So, \((-2)^2 - 4(1)(-15 + r) = 0\).

\(4 - 4(-15 + r) = 0\).

\(4 + 60 - 4r = 0\).

\(64 = 4r\).

\(r = 16\).