To find the value of \(k\) for which the line \(y = 2x + k\) is a tangent to the curve \(y = 2x^2 - 6x + 5\), we equate the two equations:

\(2x^2 - 6x + 5 = 2x + k\)

Rearrange to form a quadratic equation:

\(2x^2 - 8x + 5 - k = 0\)

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

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

Here, \(a = 2\), \(b = -8\), and \(c = 5 - k\).

Calculate the discriminant:

\((-8)^2 - 4(2)(5 - k) = 0\)

\(64 - 8(5 - k) = 0\)

\(64 - 40 + 8k = 0\)

\(24 + 8k = 0\)

\(8k = -24\)

\(k = -3\)

Thus, the value of \(k\) is \(-3\).

To show that \(4c = m^2 - 12m + 16\), we start by equating the line and the curve: \(mx + c = 6x - x^2 - 5\).

Rearrange to form a quadratic equation: \(x^2 - (6-m)x + (c+5) = 0\).

For the line to be tangent to the curve, the discriminant of this quadratic must be zero: \(b^2 - 4ac = 0\).

Here, \(a = 1\), \(b = -(6-m)\), and \(c = c+5\).

Calculate the discriminant: \((6-m)^2 - 4(1)(c+5) = 0\).

Expand and simplify: \(36 - 12m + m^2 - 4c - 20 = 0\).

Rearrange to find \(4c = m^2 - 12m + 16\).

To find the point of tangency, equate the line and curve equations:

\(kx - k = -\frac{1}{2x}\)

Multiply through by \(2x\) to clear the fraction:

\(2kx^2 - 2kx = -1\)

Rearrange to form a quadratic equation:

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

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

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

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

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

\(4k^2 - 8k = 0\)

Factor out \(4k\):

\(4k(k - 2) = 0\)

Since \(k\) is positive, \(k = 2\).

Substitute \(k = 2\) back into the line equation:

\(y = 2x - 2\)

Equate with the curve equation:

\(2x - 2 = -\frac{1}{2x}\)

Multiply through by \(2x\):

\(4x^2 - 4x + 1 = 0\)

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

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

Substitute \(x = \frac{1}{2}\) into the line equation:

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

The point of tangency is \(\left( \frac{1}{2}, -1 \right)\).

To find the values of m for which the line y = mx is a tangent to the curve y = 2x^2 - 4x + 8, we equate the two equations:

\(mx = 2x^2 - 4x + 8\)

Rearrange to form a quadratic equation:

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

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

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

Here, \(a = 2\), \(b = -(4 + m)\), \(c = 8\).

Substitute into the discriminant formula:

\((-(4 + m))^2 - 4 \times 2 \times 8 = 0\)

\((4 + m)^2 - 64 = 0\)

\((4 + m)^2 = 64\)

Take the square root of both sides:

\(4 + m = 8\) or \(4 + m = -8\)

Solve for \(m\):

\(m = 4\) or \(m = -12\)

To find when the line does not intersect the curve, set the equations equal to each other:

\(x^2 - 4x + c = 2x - 7\)

Rearrange to form a quadratic equation:

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

For the line not to intersect the curve, the quadratic equation must have no real roots. This occurs when the discriminant is less than zero:

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

Here, \(a = 1\), \(b = -6\), and \(c = c + 7\). Calculate the discriminant:

\((-6)^2 - 4(1)(c + 7) < 0\)

\(36 - 4(c + 7) < 0\)

\(36 - 4c - 28 < 0\)

\(8 - 4c < 0\)

\(8 < 4c\)

\(c > 2\)