(a) Partial Fraction Form:

By division or otherwise:
\( f(x) = \frac{2x – 1}{x + 2} = 2 – \frac{5}{x + 2} \)
Answer: \( \boxed{2 – \frac{5}{x + 2}} \)

(b) Derivative and Positivity:

Differentiate \( f(x) \):
\( f'(x) = \frac{5}{(x + 2)^2} \)
Since \( (x + 2)^2 > 0 \) for all \( x \in D \), \( f'(x) > 0 \).
Conclusion: \( \boxed{f'(x) > 0 \text{ on } D} \)

(c) Range of \( f \):

Evaluate \( f \) at endpoints of \( D \):
\( f(-1) = \frac{2(-1) – 1}{-1 + 2} = -3 \)
\( f(8) = \frac{2(8) – 1}{8 + 2} = \frac{15}{10} = \frac{3}{2} \)
Since \( f \) is increasing, the range is \( \boxed{\left[ -3, \frac{3}{2} \right]} \)

(d) Inverse Function and Graphs:

(i) Let \( y = f(x) \):
\( y = \frac{2x – 1}{x + 2} \)
Swap \( x \) and \( y \), then solve for \( y \):
\( x = \frac{2y – 1}{y + 2} \)
\( xy + 2x = 2y – 1 \)
\( y(x – 2) = -2x – 1 \)
\( y = \frac{2x + 1}{2 – x} \)
Answer: \( \boxed{f^{-1}(x) = \frac{2x + 1}{2 – x}} \)

(ii) & (iii) Graphs:
Key Points:
– \( y = f(x) \) has x-intercept at \( \left( \frac{1}{2}, 0 \right) \) and y-intercept at \( \left( 0, -\frac{1}{2} \right) \).
– \( y = f^{-1}(x) \) is the reflection of \( y = f(x) \) across \( y = x \).

(e) Absolute Value Transformation:

(i) Graph of \( y = f(|x|) \):
Behavior:
– For \( x \geq 0 \), same as \( y = f(x) \).
– For \( x < 0 \), symmetric about the y-axis.

(ii) Solve \( f(|x|) = -\frac{1}{4} \):
Let \( |x| = t \), then \( \frac{2t – 1}{t + 2} = -\frac{1}{4} \).
Cross-multiply: \( 8t – 4 = -t – 2 \)
\( 9t = 2 \) ⇒ \( t = \frac{2}{9} \)
Thus, \( x = \boxed{\frac{2}{9}} \) or \( x = \boxed{-\frac{2}{9}} \).