Let's solve the given problems step-by-step based on the image provided.

Given:

1. $f(x) = \frac{5}{x - 1}$
2. $g(x) = 4 - 2x$

Part (a): $f(f^{-1}(-2))$

Steps:

1. Find $f^{-1}(x)$: To find the inverse, let $y = f(x)$: $y = \frac{5}{x - 1}$ Solve for $x$ in terms of $y$: $y(x - 1) = 5 \quad \implies \quad x - 1 = \frac{5}{y} \quad \implies \quad x = \frac{5}{y} + 1$ Thus, $f^{-1}(x) = \frac{5}{x} + 1$.

2. Evaluate $f^{-1}(-2)$: Substitute $x = -2$ into $f^{-1}(x)$: $f^{-1}(-2) = \frac{5}{-2} + 1 = -\frac{5}{2} + 1 = -\frac{3}{2}.$

3. Find $f(f^{-1}(-2))$: Substitute $x = -\frac{3}{2}$ into $f(x)$: $f\left(-\frac{3}{2}\right) = \frac{5}{\left(-\frac{3}{2} - 1\right)} = \frac{5}{-\frac{3}{2} - \frac{2}{2}} = \frac{5}{-\frac{5}{2}} = -2.$

Answer for (a): $f(f^{-1}(-2)) = -2$

Part (b): $g(g(3))$

Steps:

1. Find $g(3)$: Substitute $x = 3$ into $g(x)$: $g(3) = 4 - 2(3) = 4 - 6 = -2.$

2. Find $g(g(3))$: Substitute $x = -2$ into $g(x)$: $g(-2) = 4 - 2(-2) = 4 + 4 = 8.$

Answer for (b): $g(g(3)) = 8$

Part (c): Simplified expression for $f(g(x))$

Steps:

1. Substitute $g(x) = 4 - 2x$ into $f(x)$: $f(g(x)) = f(4 - 2x) = \frac{5}{(4 - 2x) - 1}.$

2. Simplify the denominator: $f(g(x)) = \frac{5}{4 - 2x - 1} = \frac{5}{3 - 2x}.$

Answer for (c): $f(g(x)) = \frac{5}{3 - 2x}$

Final Answers:

1. (a): $f(f^{-1}(-2)) = -2$
2. (b): $g(g(3)) = 8$
3. (c): $f(g(x)) = \frac{5}{3 - 2x}$

Would you like me to go into further details or clarify any part? Here are 5 related questions to expand on this topic:

1. How do you verify if two functions are inverses of each other?
2. Can you compute the derivative of $f(x) = \frac{5}{x - 1}$?
3. What is the range of $f(x) = \frac{5}{x - 1}$?
4. How can the composition $f(g(x))$ be used in real-world applications?
5. What happens to $g(g(x))$ if $g(x) = ax + b$?

Tip: When finding inverse functions, always confirm that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.