To find the values of k for which the line intersects the curve at two distinct points, we set the equations equal to each other:

\(kx - 4 = x^2 - 2x\)

Rearrange to form a quadratic equation:

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

For the line to intersect the curve at two distinct points, the discriminant of this quadratic equation must be greater than zero. The discriminant \(\Delta\) is given by:

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

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

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

\(\Delta = (2 + k)^2 - 16\)

For two distinct points, \(\Delta > 0\):

\((2 + k)^2 > 16\)

Solving this inequality:

\(2 + k > 4\) or \(2 + k < -4\)

\(k > 2\) or \(k < -6\)

To find the values of k for which the line y = 4x + k does not intersect the curve y = x^2, we set the equations equal to each other:

\(x^2 = 4x + k\)

Rearrange to form a quadratic equation:

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

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

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

Here, a = 1, b = -4, and c = -k. Substitute these into the discriminant:

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

Simplify:

\(16 + 4k < 0\)

Solve for k:

\(4k < -16\)

\(k < -4\)

Thus, the set of values for k is k < -4.

To find the value of c, we need the line y = 2x + c to be tangent to the curve y² = 4x. This means they intersect at exactly one point.

Substitute y = 2x + c into y² = 4x:

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

Expand and simplify:

\(4x^2 + 4xc + c^2 = 4x\)

Rearrange to form a quadratic equation:

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

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

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

Here, \(a = 4\), \(b = 4c - 4\), \(c = c^2\).

Calculate the discriminant:

\((4c - 4)^2 - 4 \times 4 \times c^2 = 0\)

\(16c^2 - 32c + 16 - 16c^2 = 0\)

\(-32c + 16 = 0\)

Solve for c:

\(-32c = -16\)

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

To find the set of values of \(k\) for which the line \(l\) does not intersect the curve, substitute \(y = k - 2x\) from the line equation into the curve equation \(xy = 12\).

This gives:

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

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

Rearrange to form a quadratic equation:

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

For the line not to intersect the curve, the quadratic equation must have no real roots. This occurs when the discriminant \(b^2 - 4ac < 0\).

Here, \(a = 2\), \(b = -k\), and \(c = 12\).

Calculate the discriminant:

\((-k)^2 - 4(2)(12) < 0\)

\(k^2 - 96 < 0\)

Thus, \(k^2 < 96\).

Taking the square root of both sides gives:

\(-\sqrt{96} < k < \sqrt{96}\)

Therefore, the set of values of \(k\) for which the line does not intersect the curve is \(|k| < \sqrt{96}\).

Substitute \(a = -\frac{7}{2}\) into the line equation, giving \(y = x + \frac{7}{2}\).

Set the curve equal to the line: \(4x^2 - kx + \frac{1}{2}k^2 = x + \frac{7}{2}\).

Rearrange to form a quadratic equation: \(4x^2 - (k+1)x + \left(\frac{1}{2}k^2 - \frac{7}{2}\right) = 0\).

For the line to be tangent, the discriminant must be zero: \((k+1)^2 - 4 \times 4 \times \left(\frac{1}{2}k^2 - \frac{7}{2}\right) = 0\).

Simplify: \((k+1)^2 - 2k^2 + 14 = 0\).

Further simplify to: \(7k^2 - 2k - 57 = 0\).

Factor or use the quadratic formula to solve: \((k-3)(7k+19) = 0\).

Thus, \(k = 3\) or \(k = \frac{19}{7}\).