The curve $C$ has equation $y=\frac{ax^2-2ax+a-2}{x-2},\quad a\in\mathbb{R},\ a\ne 0.$

1. Apply the quotient rule to differentiate $y$ with respect to $x$.

Let $u = ax^2 - 2ax + a - 2$ and $v = x - 2$.
Then $\frac{\mathrm{d}u}{\mathrm{d}x} = 2ax - 2a$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = 1$.
Using the quotient rule $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x - 2)(2ax - 2a) - (ax^2 - 2ax + a - 2)(1)}{(x - 2)^2}$

2. Expand and simplify the numerator to get the final derivative.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2ax^2 - 2ax - 4ax + 4a - ax^2 + 2ax - a + 2}{(x - 2)^2}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ax^2 - 4ax + 3a + 2}{(x - 2)^2}$

1. Set the numerator of the derivative to zero to find the condition for stationary points.

From part (i), we have:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ax^2 - 4ax + 3a + 2}{(x-2)^2}$
For stationary points, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, which means the numerator must be zero:
$ax^2 - 4ax + 3a + 2 = 0$

2. Apply the condition for the curve to have no stationary points.

For the curve to have no stationary points, the quadratic equation must have no real roots. Therefore, the discriminant must be less than zero ($\Delta < 0$):
$b^2 - 4ac < 0$
$(-4a)^2 - 4(a)(3a + 2) < 0$

3. Solve the inequality to find the range of values for $a$.

$16a^2 - 12a^2 - 8a < 0$
$4a^2 - 8a < 0$
$4a(a - 2) < 0$
Since the parabola opens upwards and we want the region below the $a$-axis:
$0 < a < 2$

1. Substitute $a = 1$ into the curve equation and find the equations of the asymptotes.

Substituting $a = 1$ into $C$,
$y = \frac{x^2 - 2x - 1}{x - 2}$
Performing polynomial division to find the oblique asymptote:
$y = \frac{x(x - 2) - 1}{x - 2} = x - \frac{1}{x - 2}$
From this form, the vertical asymptote is $x = 2$.
As $x \to \pm\infty$, $\frac{1}{x - 2} \to 0$, so the oblique asymptote is $y = x$.

2. Calculate the coordinates of the intersections with the axes.

For the $y$-intercept, set $x = 0$:
$y = \frac{-1}{-2} = \frac{1}{2}$
The intersection is $(0, \frac{1}{2})$.

For the $x$-intercepts, set $y = 0$:
$x^2 - 2x - 1 = 0$
Using the quadratic formula:
$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$
The intersections are $(1-\sqrt{2}, 0)$ and $(1+\sqrt{2}, 0)$.

3. Determine if there are any turning points by analysing the first derivative.

Differentiating $y = x - (x-2)^{-1}$ with respect to $x$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + (x-2)^{-2} = 1 + \frac{1}{(x-2)^2}$
Since $\frac{1}{(x-2)^2} > 0$ for all $x \ne 2$, we have $\frac{\mathrm{d}y}{\mathrm{d}x} > 1 > 0$ everywhere on the curve's domain.
Therefore, $\frac{\mathrm{d}y}{\mathrm{d}x} \ne 0$ for all $x$, meaning there are no turning points.

4. Describe the sketch of the curve incorporating all calculated features.

Draw the asymptotes $x=2$ and $y=x$ as dashed lines.
Plot the $y$-intercept $(0, \frac{1}{2})$ and $x$-intercepts $(1-\sqrt{2}, 0)$ and $(1+\sqrt{2}, 0)$.
The left branch ($x < 2$) lies above the oblique asymptote (since $-\frac{1}{x-2} > 0$), passes through $(1-\sqrt{2}, 0)$ and $(0, \frac{1}{2})$, and curves upwards towards $x = 2$.
The right branch ($x > 2$) lies below the oblique asymptote, comes from $-\infty$ near $x = 2$, passes through $(1+\sqrt{2}, 0)$, and approaches $y=x$ from below as $x \to \infty$.

1. Rearrange the given equation to identify the suitable curve to sketch alongside $C$.

$x^3-2x^2-x = k(x-2)$
Factorising out $x$ on the left-hand side:
$x(x^2-2x-1) = k(x-2)$
Dividing both sides by $x(x-2)$ (noting that $x=0$ and $x=2$ are not solutions):
$\frac{x^2-2x-1}{x-2} = \frac{k}{x}$
The left-hand side is the equation of the curve $C$ for $a=1$. Thus, the roots of the equation correspond to the $x$-coordinates of the points of intersection between the curve $C$ and the curve $y = \frac{k}{x}$.

2. Determine the number of intersections by visualising the sketch of $y = \frac{k}{x}$.

Sketch the curve $y = \frac{k}{x}$ on the same diagram from (iii). Since $k>0$, the hyperbola has branches in the first ($x>0, y>0$) and third ($x<0, y<0$) quadrants.

From the visual behaviour of the graphs:

• For $x < 0$: The branch of $y = \frac{k}{x}$ in the third quadrant intersects the left portion of $C$ exactly once, yielding $1$ negative root.
• For $x > 0$: The branch of $y = \frac{k}{x}$ in the first quadrant intersects the left branch of $C$ ($0 < x < 2$) exactly once, and it also intersects the right branch of $C$ ($x > 2$) exactly once. This yields a total of $2$ positive roots.