(i) To find when the curve \(y = kx^2 + 1\) and the line \(y = kx\) have no common points, set the equations equal: \(kx^2 + 1 = kx\).

Rearrange to form a quadratic equation: \(kx^2 - kx + 1 = 0\).

For no common points, the discriminant must be less than zero: \(b^2 - 4ac < 0\).

Here, \(a = k\), \(b = -k\), \(c = 1\). So, \((-k)^2 - 4(k)(1) < 0\).

Simplify: \(k^2 - 4k < 0\).

Factor: \(k(k - 4) < 0\).

The solution to this inequality is \(0 < k < 4\).

(ii) For the line to be tangent to the curve, the discriminant must be zero: \(k^2 - 4k = 0\).

Factor: \(k(k - 4) = 0\).

Since \(k \neq 0\), \(k = 4\).

Substitute \(k = 4\) into the quadratic equation: \(4x^2 - 4x + 1 = 0\).

Factor: \((2x - 1)^2 = 0\).

Solve for \(x\): \(x = \frac{1}{2}\).

Substitute \(x = \frac{1}{2}\) into \(y = 4x\): \(y = 4 \times \frac{1}{2} = 2\).

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

To find the intersection points, set the equations equal: \(\frac{9}{2-x} = x + k\).

Multiply both sides by \(2-x\) to eliminate the fraction: \(9 = (x + k)(2-x)\).

Expand and rearrange: \(x^2 - 2x + kx - 2k + 9 = 0\).

This simplifies to \(x^2 + (k-2)x + (9-2k) = 0\).

For two distinct points, the discriminant \(b^2 - 4ac\) must be greater than zero.

Here, \(a = 1\), \(b = k-2\), \(c = 9-2k\).

Calculate the discriminant: \((k-2)^2 - 4(1)(9-2k) > 0\).

Simplify: \(k^2 - 4k + 4 - 36 + 8k > 0\).

Further simplify: \(k^2 + 4k - 32 > 0\).

Factorize: \((k+8)(k-4) > 0\).

The solution to this inequality is \(k < -8\) or \(k > 4\).

To find the values of \(k\) for which the line \(y + kx = 12\) is a tangent to the curve \(y = 2x^2 - 8x + 14\), we first express the line equation in terms of \(y\):

\(y = -kx + 12\)

Set the expressions for \(y\) equal to each other:

\(2x^2 - 8x + 14 = -kx + 12\)

Rearrange to form a quadratic equation:

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

For the line to be tangent to the curve, the quadratic must have exactly one solution, which occurs when the discriminant is zero:

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

Here, \(a = 2\), \(b = k-8\), and \(c = 2\). Substitute these into the discriminant formula:

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

Simplify:

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

Take the square root of both sides:

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

Solve for \(k\):

\(k = 12 \quad \text{or} \quad k = 4\)

To find the set of values of k for which the line does not intersect the curve, we substitute y from the line equation into the curve equation:

\(y = \frac{x + k}{2}\)

Substitute into the curve equation:

\(\frac{x + k}{2} = x^2 - 4x + 7\)

Multiply through by 2 to eliminate the fraction:

\(x + k = 2x^2 - 8x + 14\)

Rearrange to form a quadratic equation:

\(2x^2 - 9x + 14 - k = 0\)

For the line not to intersect the curve, the discriminant of this quadratic must be less than zero:

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

Here, a = 2, b = -9, c = 14 - k. Calculate the discriminant:

\((-9)^2 - 4 \times 2 \times (14 - k) < 0\)

\(81 - 8(14 - k) < 0\)

Simplify:

\(81 - 112 + 8k < 0\)

\(8k < 31\)

\(k < \frac{31}{8}\)

Thus, the set of values of k for which the line does not intersect the curve is:

\(k < 3.875\)

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

Substitute these values into the discriminant condition:

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

\(k^2 - 64 < 0\)

\(k^2 < 64\)

This implies \(-8 < k < 8\).