Let's go through each question one by one and solve them.

1. Given $f(x) = 8x + 5$ and $g(x) = x + 2$, if $(f \circ g)(x) = 37$, find $x$.

   • Start by calculating $f(g(x))$: $f(g(x)) = f(x + 2) = 8(x + 2) + 5 = 8x + 16 + 5 = 8x + 21$
   • Set $8x + 21 = 37$ and solve for $x$: $8x = 16 \Rightarrow x = 2$

2. Given $g(x) = x - 3$ and $(g \circ f)(x) = 2x - 2$, find $f(x)$.

   • Start by expressing $(g \circ f)(x) = g(f(x))$: $g(f(x)) = f(x) - 3$
   • Since $(g \circ f)(x) = 2x - 2$, set up the equation: $f(x) - 3 = 2x - 2$
   • Solve for $f(x)$: $f(x) = 2x + 1$

3. If $f(x) = x + 4$, $g(x) = x^2 + 1$, and $h(x) = 3x$, find $(h \circ g \circ f)(x)$.

   • First, calculate $g(f(x))$: $f(x) = x + 4 \Rightarrow g(f(x)) = (x + 4)^2 + 1 = x^2 + 8x + 16 + 1 = x^2 + 8x + 17$
   • Now calculate $h(g(f(x)))$: $h(g(f(x))) = h(x^2 + 8x + 17) = 3(x^2 + 8x + 17) = 3x^2 + 24x + 51$

4. If $f(x) = \frac{3x - 1}{4x - 2}$, find $f^{-1}(0)$.

   • Set $f(x) = 0$: $\frac{3x - 1}{4x - 2} = 0$
   • Solve for $x$ when the numerator is zero: $3x - 1 = 0 \Rightarrow x = \frac{1}{3}$
   • Therefore, $f^{-1}(0) = \frac{1}{3}$.

5. Given $f(x) = x - 4$ and $g(x) = 3x + 2$, find $(f \circ g)^{-1}(x)$.

   • First, calculate $(f \circ g)(x)$: $f(g(x)) = f(3x + 2) = 3x + 2 - 4 = 3x - 2$
   • Now, to find $(f \circ g)^{-1}(x)$, set $y = 3x - 2$ and solve for $x$: $y + 2 = 3x \Rightarrow x = \frac{y + 2}{3}$
   • So, $(f \circ g)^{-1}(x) = \frac{x + 2}{3}$.

Would you like further explanations or details on any of these steps?

Here are some questions to deepen your understanding:

1. What would happen if $f(x)$ and $g(x)$ were switched in Question 1?
2. How do you find the inverse of a function in general?
3. What is the significance of function composition in mathematics?
4. How can we verify that our answer to an inverse function is correct?
5. What are some practical applications of function composition?

Tip: When dealing with composite functions, always work step-by-step from the innermost function to the outermost for clarity and accuracy.