The equation of the line is given by y = mx - 2.

To find when the line is tangent to the curve y = x^2 - 2x + 7, equate the two equations:

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

Rearrange to form a quadratic equation:

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

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

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

Substitute into the discriminant formula:

\((2 + m)^2 - 4 \times 1 \times 9 = 0\)

Simplify and solve for m:

\((2 + m)^2 = 36\)

\(2 + m = 6 \quad \text{or} \quad 2 + m = -6\)

\(m = 4 \quad \text{or} \quad m = -8\)

For m = 4, substitute back to find x:

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

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

\(x = 3\)

Substitute x = 3 into the line equation to find y:

\(y = 4 \times 3 - 2 = 10\)

Coordinates are (3, 10).

For m = -8, substitute back to find x:

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

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

\(x = -3\)

Substitute x = -3 into the line equation to find y:

\(y = -8 \times (-3) - 2 = 22\)

Coordinates are (-3, 22).

For the quadratic equation \(ax^2 + bx + c = 0\) to have no real roots, the discriminant must be less than zero.

The discriminant \(\Delta\) is given by \(b^2 - 4ac\).

Here, \(a = 4\), \(b = -24\), and \(c = p\).

So, \(\Delta = (-24)^2 - 4 \times 4 \times p\).

\(\Delta = 576 - 16p\).

For no real roots, \(576 - 16p < 0\).

Solving for \(p\):

\(576 < 16p\)

\(p > 36\)

(i) To find the value of \(c\), we need the line \(4y = x + c\) to be tangent to the curve \(y^2 = x + 3\). Substitute \(x = 4y - c\) into the curve equation:

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

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

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

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

Here, \(a = 1\), \(b = -4\), \(c = c - 3\). So:

\((-4)^2 - 4(1)(c - 3) = 0\)

\(16 - 4c + 12 = 0\)

\(28 - 4c = 0\)

\(4c = 28\)

\(c = 7\)

(ii) With \(c = 7\), substitute back to find \(P\):

\(y^2 = 4y - 7 + 3\)

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

\((y - 2)^2 = 0\)

\(y = 2\)

Substitute \(y = 2\) into \(x = 4y - c\):

\(x = 4(2) - 7 = 1\)

Coordinates of \(P\) are \((1, 2)\).

(i) To show that the curve and the line meet for all values of \(k\), equate the equations:

\(2x^2 - 3x + 1 = kx + k^2\)

Rearrange to form a quadratic equation:

\(2x^2 - (3+k)x + (1-k^2) = 0\)

For the curve and line to meet, the discriminant \(b^2 - 4ac\) must be non-negative:

\((3+k)^2 - 4 \times 2 \times (1-k^2) \geq 0\)

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

\(9k^2 + 6k + 1 \geq 0\)

\((3k + 1)^2 \geq 0\)

This is always true, hence the curve and line meet for all values of \(k\).

(ii) For the line to be tangent to the curve, the discriminant must be zero:

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

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

\(9k^2 + 6k + 1 = 0\)

Solving for \(k\), we find \(k = -\frac{1}{3}\).

Substitute \(k = -\frac{1}{3}\) into the line equation:

\(y = -\frac{1}{3}x + \left(-\frac{1}{3}\right)^2 = -\frac{1}{3}x + \frac{1}{9}\)

Equate with the curve equation:

\(2x^2 - 3x + 1 = -\frac{1}{3}x + \frac{1}{9}\)

\(2x^2 - \frac{8}{3}x + \frac{8}{9} = 0\)

Solving gives \(x = \frac{2}{3}\).

Substitute \(x = \frac{2}{3}\) back into the line equation to find \(y\):

\(y = -\frac{1}{3} \times \frac{2}{3} + \frac{1}{9} = -\frac{1}{9}\)

Thus, the point of tangency is \(\left( \frac{2}{3}, -\frac{1}{9} \right)\).

To find the set of values of \(k\) for which the line does not meet the curve, we need to solve the system of equations:

1. \(y = 2x + \frac{12}{x}\)

2. \(y + x = k\)

Substitute \(y = k - x\) from the second equation into the first equation:

\(k - x = 2x + \frac{12}{x}\)

Rearrange to form a quadratic equation:

\(2x + \frac{12}{x} = k - x\)

\(2x^2 + 12 = kx - x^2\)

\(3x^2 - kx + 12 = 0\)

For the line not to meet the curve, the quadratic equation must have no real solutions. This occurs when the discriminant \(b^2 - 4ac\) is less than zero.

Here, \(a = 3\), \(b = -k\), \(c = 12\).

The discriminant is:

\((-k)^2 - 4 \times 3 \times 12\)

\(k^2 - 144 \lt 0\)

Solving \(k^2 - 144 \lt 0\) gives:

\(-12 \lt k \lt 12\)