在同一幅图上画出 $y = 1 + \frac{a - 2}{x - a}$ 和 $y = -\frac{1}{a}x + \frac{2}{a}$ 的图像，其中 $a$ 是一个正常数且 $1 < a < 2$，并清晰标出所有渐近线和与坐标轴的交点。

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:
$0 < x < \frac{1}{2} \quad \text{or} \quad x > \frac{1}{a}$