(i) The coordinates of \(P\) are \((t, 5t)\) since it lies on the line \(y = 5x\). The coordinates of \(Q\) are \((t, t(9 - t^2))\) since it lies on the curve \(y = x(9 - x^2)\). The length of \(PQ\) is the difference in the \(y\)-coordinates of \(Q\) and \(P\):

\(PQ = t(9 - t^2) - 5t = 9t - t^3 - 5t = 4t - t^3\)

(ii) To find the maximum value of \(PQ\), differentiate \(PQ = 4t - t^3\) with respect to \(t\):

\(\frac{d(PQ)}{dt} = 4 - 3t^2\)

Set the derivative to zero to find critical points:

\(4 - 3t^2 = 0\)

\(3t^2 = 4\)

\(t^2 = \frac{4}{3}\)

\(t = \frac{2}{\sqrt{3}}\)

Substitute \(t = \frac{2}{\sqrt{3}}\) back into \(PQ = 4t - t^3\) to find the maximum length:

\(PQ = 4\left(\frac{2}{\sqrt{3}}\right) - \left(\frac{2}{\sqrt{3}}\right)^3\)

\(PQ = \frac{8}{\sqrt{3}} - \frac{8}{3\sqrt{3}}\)

\(PQ = \frac{24}{3\sqrt{3}} - \frac{8}{3\sqrt{3}}\)

\(PQ = \frac{16}{3\sqrt{3}}\)

Alternatively, \(PQ = \frac{16\sqrt{3}}{9}\).

(i) To find the stationary point B, we set the derivative of the curve to zero: \(\frac{dy}{dx} = 2x - 4 = 0\). Solving gives \(x = 2\). Substituting \(x = 2\) into the curve equation gives \(y = 3\). Thus, B is (2, 3).

The midpoint of AB is \(\left( \frac{4+2}{2}, \frac{7+3}{2} \right) = (3, 5)\).

Since L passes through this midpoint, substitute (3, 5) into \(y = mx - 2\):

\(5 = 3m - 2\)

Solving for \(m\), we get \(m = \frac{7}{3}\).

(ii) The line L does not meet the curve if the quadratic equation formed by substituting \(y = mx - 2\) into the curve equation has no real solutions. This occurs when the discriminant is less than zero.

Substitute \(y = mx - 2\) into \(y = x^2 - 4x + 7\):

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

Rearrange to form a quadratic: \(x^2 - (4+m)x + 9 = 0\)

The discriminant \(b^2 - 4ac\) is \((m+4)^2 - 36\).

Set \((m+4)^2 - 36 < 0\) and solve:

\((m+4)^2 < 36\)

\(-6 < m+4 < 6\)

\(-10 < m < 2\)

(i) To find the points of intersection, set the equations equal: \(x^2 - x + 3 = 3x + a\). Rearrange to get \(x^2 - 4x + (3 - a) = 0\).

(ii) Substitute \(x = -1\) into the equation \(x^2 - 4x + (3 - a) = 0\):

\((-1)^2 - 4(-1) + (3 - a) = 0\)

\(1 + 4 + 3 - a = 0\)

\(8 - a = 0 \Rightarrow a = 8\)

Substitute \(a = 8\) back into the equation:

\(x^2 - 4x - 5 = 0\)

Factorize: \((x - 5)(x + 1) = 0\)

The other \(x\)-coordinate is \(x = 5\).

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

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

\((-4)^2 - 4(1)(3 - a) = 0\)

\(16 - 4(3 - a) = 0\)

\(16 - 12 + 4a = 0\)

\(4a = -4 \Rightarrow a = -1\)

Substitute \(a = -1\) into \(x^2 - 4x + 4 = 0\):

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

Substitute \(x = 2\) into \(y = x^2 - x + 3\):

\(y = 2^2 - 2 + 3 = 5\)

The coordinates of \(P\) are \((2, 5)\).

(i) Equate the line and curve equations: \(-2x + k = \frac{2}{x - 3}\).

Multiply through by \(x - 3\) to clear the fraction: \(-2x(x - 3) + k(x - 3) = 2\).

Simplify to get \(2x^2 - (6 + k)x + (2 + 3k) = 0\).

(ii) For the line to be tangent to the curve, the quadratic equation must have a repeated root, so the discriminant \(b^2 - 4ac = 0\).

Here, \(a = 2\), \(b = -(6 + k)\), \(c = 2 + 3k\).

Calculate the discriminant: \((6 + k)^2 - 4(2)(2 + 3k) = 0\).

Simplify: \(k^2 - 12k + 20 = 0\).

Factorize: \((k - 10)(k - 2) = 0\).

Thus, \(k = 2\) or \(k = 10\).

(iii) For \(k = 2\), substitute into the quadratic: \(2(x - 2)^2 = 0\).

\(x = 2\), \(y = -2\), so \(A = (2, -2)\).

For \(k = 10\), substitute into the quadratic: \(2(x - 4)^2 = 0\).

\(x = 4\), \(y = 2\), so \(B = (4, 2)\).

Find the equation of line \(AB\):

Slope \(m = \frac{2 - (-2)}{4 - 2} = 2\).

Using point-slope form: \(y + 2 = 2(x - 2)\) or \(y = 2x - 6\).

(i) To find the \(x\)-coordinates of \(A\) and \(B\), set the equations equal: \(2x^5 + 3x^3 = 2x\).

Rearrange to: \(2x^5 + 3x^3 - 2x = 0\).

Factor out \(x\): \(x(2x^4 + 3x^2 - 2) = 0\).

Thus, \(2x^4 + 3x^2 - 2 = 0\) is satisfied by the \(x\)-coordinates of \(A\) and \(B\).

(ii) Solve \(2x^4 + 3x^2 - 2 = 0\) by substituting \(y = x^2\), giving \(2y^2 + 3y - 2 = 0\).

Factorize: \((y + 2)(2y - 1) = 0\).

Thus, \(y = -2\) or \(y = \frac{1}{2}\).

Since \(y = x^2\), \(x^2 = \frac{1}{2}\) (ignore \(y = -2\) as \(x^2\) cannot be negative).

So, \(x = \pm \frac{1}{\sqrt{2}}\).

For \(x = \frac{1}{\sqrt{2}}\), \(y = 2x = \frac{2}{\sqrt{2}}\).

For \(x = -\frac{1}{\sqrt{2}}\), \(y = 2x = -\frac{2}{\sqrt{2}}\).

Thus, the coordinates are \(\left( \frac{1}{\sqrt{2}}, \frac{2}{\sqrt{2}} \right)\) and \(\left( -\frac{1}{\sqrt{2}}, -\frac{2}{\sqrt{2}} \right)\).