Sketch the graphs of $y = 1 + \frac{a - 2}{x - a}$, and $y = -\frac{1}{a}x + \frac{2}{a}$ on a single diagram, where $a$ is a positive constant and $1 < a < 2$, showing all asymptotes and axial intercepts clearly.

1. Identify the points of intersection of the curve and the line from their axial intercepts.

From the prior sketch, the axial intercepts of the curve $y = 1 + \frac{a - 2}{x - a}$ are found at $x=0 \implies y = 1 + \frac{a-2}{-a} = \frac{2}{a}$ and $y=0 \implies \frac{a-2}{x-a} = -1 \implies x=2$.
Similarly, the axial intercepts of the line $y = -\frac{1}{a}x + \frac{2}{a}$ are at $x=0 \implies y = \frac{2}{a}$ and $y=0 \implies x=2$.
Thus, the curve and the line intersect exactly at $(0, \frac{2}{a})$ and $(2, 0)$.

2. Determine the range of $x$ where the curve is above the line.

The inequality $1 + \frac{a - 2}{x - a} > -\frac{1}{a}x + \frac{2}{a}$ means we require the curve to be strictly above the line.
Observing the relative positions of the graphs and considering the vertical asymptote is at $x = a$ (where $1 < a < 2$), the curve is above the line between the $y$-intercept and the vertical asymptote, and also to the right of the $x$-intercept.
Therefore, the solution is $0 < x < a$ or $x > 2$.

1. Determine the solution for part (i) to use in the current part

To answer the target, we first establish the solution to part (i).
Find the $x$-coordinates of the intersections of the curve and the line:
$1 + \frac{a-2}{x-a} = -\frac{1}{a}x + \frac{2}{a}$
$\frac{x-a+a-2}{x-a} = \frac{2-x}{a}$
$\frac{x-2}{x-a} = \frac{2-x}{a}$
$(x-2) \left[ \frac{1}{x-a} + \frac{1}{a} \right] = 0$
$(x-2) \left[ \frac{a+x-a}{a(x-a)} \right] = 0$
$\frac{x(x-2)}{a(x-a)} = 0 \implies x = 0 \text{ or } x = 2$
Using the properties of the graphs, the curve lies above the line for the intervals:
$0 < x < a \quad \text{or} \quad x > 2$

2. Manipulate the target inequality to match the form in part (i)

Rearrange the inequality given in (ii) to reveal its relation to (i).
Factorize $x$ in the numerator of the rational expression and then divide both the numerator and denominator by $x$:
$1 + \frac{x(a - 2)}{x\left(\frac{1}{x} - a\right)} > -\frac{1}{a}\left(\frac{1}{x}\right) + \frac{2}{a}$
$1 + \frac{a - 2}{\frac{1}{x} - a} > -\frac{1}{a}\left(\frac{1}{x}\right) + \frac{2}{a}$
This is identical to the inequality from part (i), but with $x$ replaced by $\frac{1}{x}$.

3. Apply the substitution to the solution from part (i)

Substitute $\frac{1}{x}$ into the solution bounds found in (i):
$0 < \frac{1}{x} < a \quad \text{or} \quad \frac{1}{x} > 2$

4. Solve the reciprocal inequalities to find the final bounds for $x$

Solve these inequalities for $x$.
For $0 < \frac{1}{x} < a$:
Since $a > 0$ and $\frac{1}{x} > 0$, we know $x > 0$. Taking the reciprocal reverses the inequality sign:
$x > \frac{1}{a}$

For $\frac{1}{x} > 2$:
Similarly, $\frac{1}{x}$ is positive, meaning $x > 0$. Taking the reciprocal gives:
$x < \frac{1}{2}$
Thus, $0 < x < \frac{1}{2}$.

Combining these, the final solution is:
$0 < x < \frac{1}{2} \quad \text{or} \quad x > \frac{1}{a}$