First, find the derivative of the curve \(y = 2x^3 - 5x^2 - 3x + c\):

\(\frac{dy}{dx} = 6x^2 - 10x - 3\).

Evaluate the derivative at \(x = 2\):

\(\frac{dy}{dx} = 6(2)^2 - 10(2) - 3 = 24 - 20 - 3 = 1\).

Since the line is tangent to the curve, the slope \(a\) of the line is equal to the derivative at \(x = 2\):

\(a = 1\).

Substitute \(x = 2\) and \(y = 6\) into the line equation \(y = ax + b\):

\(6 = 2(1) + b \Rightarrow b = 4\).

Substitute \(x = 2\) and \(y = 6\) into the curve equation \(y = 2x^3 - 5x^2 - 3x + c\):

\(6 = 2(2)^3 - 5(2)^2 - 3(2) + c\).

Simplify to find \(c\):

\(6 = 16 - 20 - 6 + c \Rightarrow c = 16\).

(i) To find \(\frac{dy}{dx}\), differentiate \(y = (2x - 1)^{-1} + 2x\).

Using the chain rule, \(\frac{d}{dx}((2x - 1)^{-1}) = -1(2x - 1)^{-2} \cdot 2 = -2(2x - 1)^{-2}\).

The derivative of \(2x\) is \(2\).

Thus, \(\frac{dy}{dx} = -2(2x - 1)^{-2} + 2\).

For \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = -2(2x - 1)^{-2} + 2\).

Using the chain rule again, \(\frac{d}{dx}(-2(2x - 1)^{-2}) = -2 \cdot -2(2x - 1)^{-3} \cdot 2 = 8(2x - 1)^{-3}\).

Thus, \(\frac{d^2y}{dx^2} = 8(2x - 1)^{-3}\).

(ii) To find stationary points, set \(\frac{dy}{dx} = 0\).

\(-2(2x - 1)^{-2} + 2 = 0\)

\(-2(2x - 1)^{-2} = -2\)

\((2x - 1)^{-2} = 1\)

\((2x - 1)^2 = 1\)

\(2x - 1 = \pm 1\)

\(2x = 2 \Rightarrow x = 1\)

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

For \(x = 0\), \(\frac{d^2y}{dx^2} = 8(2 \cdot 0 - 1)^{-3} = -8\), indicating a maximum.

For \(x = 1\), \(\frac{d^2y}{dx^2} = 8(2 \cdot 1 - 1)^{-3} = 8\), indicating a minimum.

(i) To find stationary points, we set the derivative equal to zero. The derivative is \(\frac{dy}{dx} = 3x^2 - 4x + 5\). The discriminant of this quadratic is \(b^2 - 4ac = (-4)^2 - 4 \times 3 \times 5 = 16 - 60 = -44\), which is negative. Therefore, there are no real roots, and the curve has no stationary points.

(ii) The gradient \(m = 3x^2 - 4x + 5\). To find the stationary value of \(m\), we differentiate \(m\) with respect to \(x\): \(\frac{dm}{dx} = 6x - 4\). Setting \(\frac{dm}{dx} = 0\) gives \(6x - 4 = 0\), so \(x = \frac{2}{3}\). Substituting \(x = \frac{2}{3}\) into \(m = 3x^2 - 4x + 5\), we get \(m = 3 \left( \frac{2}{3} \right)^2 - 4 \left( \frac{2}{3} \right) + 5 = \frac{11}{3}\).

To determine the nature, we find \(\frac{d^2m}{dx^2} = 6\), which is positive, indicating a minimum.

First, find the derivative \(f'(x)\):

\(f(x) = \frac{8}{x-2} + 2\)

\(f'(x) = \frac{-8}{(x-2)^2}\)

Next, find the inverse function \(f^{-1}(x)\):

Let \(y = \frac{8}{x-2} + 2\)

\(y - 2 = \frac{8}{x-2}\)

\(x - 2 = \frac{8}{y-2}\)

\(f^{-1}(x) = \frac{8}{x-2} + 2\)

Substitute into the inequality:

\(6f'(x) + 2f^{-1}(x) - 5 < 0\)

\(6 \left( \frac{-8}{(x-2)^2} \right) + 2 \left( \frac{8}{x-2} + 2 \right) - 5 < 0\)

\(\frac{-48}{(x-2)^2} + \frac{16}{x-2} + 4 - 5 < 0\)

\(\frac{-48}{(x-2)^2} + \frac{16}{x-2} - 1 < 0\)

Multiply through by \((x-2)^2\):

\(-48 + 16(x-2) - (x-2)^2 < 0\)

\(-48 + 16x - 32 - (x^2 - 4x + 4) < 0\)

\(-48 + 16x - 32 - x^2 + 4x - 4 < 0\)

\(x^2 - 20x + 84 < 0\)

Factor the quadratic:

\((x-6)(x-14) < 0\)

Find the intervals where the inequality holds:

\(2 < x < 6, \; x > 14\)

(i) To find the stationary points, set \(\frac{dy}{dx} = 0\). First, find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = x - 6x^{\frac{1}{2}} + 8\).

Set \(\frac{dy}{dx} = 0\):

\(x - 6x^{\frac{1}{2}} + 8 = 0\).

Let \(u = x^{\frac{1}{2}}\), then \(u^2 = x\). The equation becomes:

\(u^2 - 6u + 8 = 0\).

Solving the quadratic equation gives \(u = 4\) or \(u = 2\).

Thus, \(x = 16\) or \(x = 4\).

(ii) Differentiate \(\frac{dy}{dx} = x - 6x^{\frac{1}{2}} + 8\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 1 - 3x^{-\frac{1}{2}}\).

(iii) Evaluate \(\frac{d^2y}{dx^2}\) at the stationary points:

When \(x = 16\), \(\frac{d^2y}{dx^2} = 1 - 3 \times 16^{-\frac{1}{2}} = \frac{1}{4} > 0\), indicating a minimum point.

When \(x = 4\), \(\frac{d^2y}{dx^2} = 1 - 3 \times 4^{-\frac{1}{2}} = -\frac{1}{2} < 0\), indicating a maximum point.