To find the values of \(b\) for which the line meets the curve, set the equations equal to each other:

\(x + 1 = x^2 + bx + 5\)

Rearrange to form a quadratic equation:

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

For the line to meet the curve, the quadratic must have real roots. This requires the discriminant to be non-negative:

\((b-1)^2 - 4 \times 1 \times 4 \geq 0\)

\((b-1)^2 - 16 \geq 0\)

\((b-1)^2 \geq 16\)

Taking square roots gives:

\(b-1 \geq 4\) or \(b-1 \leq -4\)

Solving these inequalities:

\(b \geq 5\) or \(b \leq -3\)

Thus, the set of values for \(b\) is \(b \geq 5\) or \(b \leq -3\).

(i) To find when the curve \(y = x^2 - 6x + k\) lies entirely above the \(x\)-axis, the quadratic must have no real roots. This occurs when the discriminant \(b^2 - 4ac < 0\). Here, \(a = 1\), \(b = -6\), and \(c = k\). The discriminant is \((-6)^2 - 4(1)(k) = 36 - 4k\). For the curve to lie above the \(x\)-axis, \(36 - 4k < 0\), which simplifies to \(k > 9\).

(ii) For the line \(y + 2x = 7\) to be tangent to the curve, it must touch the curve at exactly one point. Substitute \(y = 7 - 2x\) into the curve equation: \(7 - 2x = x^2 - 6x + k\). Rearrange to form a quadratic: \(x^2 - 4x + (k - 7) = 0\). For tangency, the discriminant must be zero: \((-4)^2 - 4(1)(k - 7) = 0\). Simplifying gives \(16 - 4k + 28 = 0\), leading to \(44 = 4k\), so \(k = 11\).

To find the points of intersection, set the equations equal: \(ax + 3a = -\frac{2}{x}\).

Multiply through by \(x\) to eliminate the fraction: \(ax^2 + 3ax + 2 = 0\).

This is a quadratic equation in \(x\): \(ax^2 + 3ax + 2 = 0\).

For the line and curve to intersect at two distinct points, the discriminant of the quadratic must be greater than zero: \(b^2 - 4ac > 0\).

Here, \(a = a\), \(b = 3a\), \(c = 2\).

Calculate the discriminant: \((3a)^2 - 4(a)(2) > 0\).

Simplify: \(9a^2 - 8a > 0\).

Factorize: \(a(9a - 8) > 0\).

The critical points are \(a = 0\) and \(a = \frac{8}{9}\).

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

- For \(a < 0\), \(a(9a - 8) > 0\) is true.

- For \(0 < a < \frac{8}{9}\), \(a(9a - 8) < 0\) is false.

- For \(a > \frac{8}{9}\), \(a(9a - 8) > 0\) is true.

Thus, the set of values for \(a\) is \(a < 0\) or \(a > \frac{8}{9}\).

To find the values of \(k\) for which the quadratic equation \(2x^2 + 3kx + k = 0\) has distinct real roots, we need to ensure that the discriminant is greater than zero.

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\).

Here, \(a = 2\), \(b = 3k\), and \(c = k\).

Thus, the discriminant is:

\(\Delta = (3k)^2 - 4 \times 2 \times k = 9k^2 - 8k\).

For distinct real roots, \(\Delta > 0\):

\(9k^2 - 8k > 0\).

Factorizing gives:

\(k(9k - 8) > 0\).

The critical points are \(k = 0\) and \(k = \frac{8}{9}\).

Using a sign chart or test intervals, we find that the inequality holds for:

\(k < 0\) or \(k > \frac{8}{9}\).

Thus, the set of values of \(k\) for which the equation has distinct real roots is \(k < 0, k > \frac{8}{9}\).

To find when the curve and the line do not meet, we set the equations equal to each other:

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

Rearrange to form a quadratic equation:

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

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

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

Here, \(a = k\), \(b = -4\), and \(c = k\). Calculate the discriminant:

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

\(16 - 4k^2 < 0\)

Rearrange:

\(4k^2 > 16\)

\(k^2 > 4\)

Taking the square root of both sides gives:

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