To find the values of \(m\) for which the line and the curve do not meet, we set the equations equal to each other:

\(mx - 3 = 2x^2 + 5\)

Rearrange to form a quadratic equation:

\(2x^2 - mx + 8 = 0\)

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

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

Calculate the discriminant:

\((-m)^2 - 4(2)(8) = m^2 - 64\)

Set the discriminant less than zero:

\(m^2 - 64 < 0\)

\(m^2 < 64\)

Taking the square root of both sides gives:

\(-8 < m < 8\)

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

\(3x^2 + 2x + 4 = mx + 1\)

Rearranging gives:

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

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

\((2 - m)^2 - 4 imes 3 imes 3 > 0\)

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

\((2 - m)^2 > 36\)

Taking the square root of both sides:

\(|2 - m| > 6\)

This gives two inequalities:

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

Solving these inequalities:

1. \(2 - m > 6\) gives \(m < -4\)

2. \(2 - m < -6\) gives \(m > 8\)

Thus, the set of values of m for which the line and the curve intersect at two distinct points is:

\(m < -4\) or \(m > 8\)

To find the value of \(k\), we set the curve equation equal to the line equation since the line is tangent to the curve:

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

Rearrange to form a quadratic equation:

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

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

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

Here, \(a = 2\), \(b = k-2\), \(c = k-4\). Substitute these into the discriminant formula:

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

Simplify:

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

Expanding both sides:

\(k^2 - 4k + 4 = 8k - 32\)

Rearrange to form a quadratic equation:

\(k^2 - 12k + 36 = 0\)

Factorize:

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

Thus, \(k = 6\).

(a) To find when the line is tangent to the curve, substitute \(y = mx + c\) into \(xy = 16\):

\(x(mx + c) = 16\)

\(mx^2 + cx - 16 = 0\)

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

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

Here, \(a = m\), \(b = c\), \(c = -16\).

\(c^2 - 4(m)(-16) = 0\)

\(c^2 + 64m = 0\)

\(m = -\frac{c^2}{64}\)

(b) Substitute \(m = -4\) into the equation:

\(x(-4x + c) = 16\)

\(-4x^2 + cx - 16 = 0\)

For two distinct points, the discriminant must be positive:

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

\(c^2 - 4(-4)(-16) > 0\)

\(c^2 - 256 > 0\)

\(c^2 > 256\)

\(c > 16\) or \(c < -16\)

(i) To find the values of \(k\) for which the line and curve meet at two distinct points, equate the equations:

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

Rearrange to form a quadratic equation:

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

For two distinct points, the discriminant \(b^2 - 4ac > 0\).

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

Calculate the discriminant:

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

\(16k^2 - 8k - 8 > 0\)

Factorize or solve the inequality:

\((k - 1)(k + \frac{1}{2}) > 0\)

Thus, \(k > 1\) or \(k < -\frac{1}{2}\).

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

\(16k^2 - 8k - 8 = 0\)

Solving gives \(k = 1\) and \(k = -\frac{1}{2}\).

For \(k = 1\), the line is \(y = 3x - 2\) and the curve is \(y = \frac{3}{2}x + 1\).

Equate the tangents:

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

Solve for \(x\):

\(x = \frac{2}{3}\)

Thus, the tangents meet on the x-axis at \(x = \frac{2}{3}\).