To find the set of values of \(p\) for which the equation \(f(x) = p\) has no real roots, we consider the equation:

\(2x^2 - 6x + 5 - p = 0\)

This is a quadratic equation in the form \(ax^2 + bx + c = 0\) where \(a = 2\), \(b = -6\), and \(c = 5 - p\).

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:

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

For the quadratic equation to have no real roots, the discriminant must be less than zero:

\(\Delta < 0\)

Substituting the values of \(a\), \(b\), and \(c\) into the discriminant:

\(\Delta = (-6)^2 - 4 \times 2 \times (5 - p)\)

\(\Delta = 36 - 8(5 - p)\)

\(\Delta = 36 - 40 + 8p\)

\(\Delta = 8p - 4\)

Setting \(\Delta < 0\):

\(8p - 4 < 0\)

\(8p < 4\)

\(p < \frac{1}{2}\)

Thus, the set of values of \(p\) for which the equation \(f(x) = p\) has no real roots is \(p < \frac{1}{2}\).

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

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

Rearrange to form a quadratic equation:

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

For the quadratic to have two distinct solutions, the discriminant must be greater than zero:

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

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

\((k - 2)^2 - 4(k - 2) > 0\)

Simplify:

\((k - 2)^2 - 4(k - 2) = (k - 2)(k - 6) > 0\)

This inequality holds when:

\(k < 2 \text{ or } k > 6\)

First, rewrite the equation \(2x^2 - 10x + 8 = kx\) as a standard quadratic equation:

\(2x^2 - (10 + k)x + 8 = 0\).

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

\(b^2 - 4ac < 0\).

Here, \(a = 2\), \(b = -(10 + k)\), and \(c = 8\).

Calculate the discriminant:

\((-(10 + k))^2 - 4 \times 2 \times 8 < 0\)

\((10 + k)^2 - 64 < 0\)

\(100 + 20k + k^2 - 64 < 0\)

\(k^2 + 20k + 36 < 0\).

To find the values of \(k\), solve the inequality:

Factor the quadratic:

\((k + 18)(k + 2) < 0\).

The critical points are \(k = -18\) and \(k = -2\).

Test intervals around these points to determine where the inequality holds:

- For \(k < -18\), both factors are negative, so the product is positive.

- For \(-18 < k < -2\), one factor is negative and the other is positive, so the product is negative.

- For \(k > -2\), both factors are positive, so the product is positive.

Thus, the solution is \(-18 < k < -2\).

To find the value of \(c\) where the line is tangent to the curve, we need to set the equations equal and solve for \(x\) and \(c\).

First, equate the line and curve equations:

\(2x + c = 8 - 2x - x^2\)

Rearrange to form a quadratic equation:

\(x^2 + 4x + c - 8 = 0\)

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

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

Here, \(a = 1\), \(b = 4\), and \(c = c - 8\).

Substitute into the discriminant formula:

\(4^2 - 4(1)(c - 8) = 0\)

Simplify:

\(16 - 4c + 32 = 0\)

\(48 - 4c = 0\)

Solve for \(c\):

\(4c = 48\)

\(c = 12\)

To find the value of \(m\) and the coordinates of \(P\), we equate the line and the curve:

\(mx + 14 = \frac{12}{x} + 2\)

Multiply through by \(x\) to eliminate the fraction:

\(mx^2 + 14x = 12 + 2x\)

Rearrange to form a quadratic equation:

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

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

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

Substitute \(a = m\), \(b = 12\), \(c = -12\):

\(12^2 - 4m(-12) = 0\)

\(144 + 48m = 0\)

\(48m = -144\)

\(m = -3\)

Substitute \(m = -3\) back into the equation to find \(x\):

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

Factor or solve for \(x\):

\(x = 2\)

Substitute \(x = 2\) into the original curve equation to find \(y\):

\(y = \frac{12}{2} + 2 = 6 + 2 = 8\)

Thus, the coordinates of \(P\) are \((2, 8)\).