Let's go through the two parts of the problem.

Part (a): Finding the domain of $f(x) = 7 + 5 \arcsin(6x - 4)$

The function involves an arcsine ($\arcsin$) term, which has the domain constraint: $-1 \leq (6x - 4) \leq 1$

1. First, solve for $x$: $-1 \leq 6x - 4 \leq 1$

2. Add 4 to all sides: $3 \leq 6x \leq 5$

3. Now, divide by 6: $\frac{1}{2} \leq x \leq \frac{5}{6}$

Thus, the domain of $f(x)$ is $\left[ \frac{1}{2}, \frac{5}{6} \right]$.

Part (b): Finding the range of $f(x) = 8 + 2 \arctan(7x + 3)$

The range of the arctangent ($\arctan$) function is $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$. Since this function is modified, we need to adjust the range based on the transformations applied.

1. The arctangent term is multiplied by 2, so the range becomes: $2 \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) = \left( -\pi, \pi \right)$

2. Now, adding 8 shifts the range: $\left( 8 - \pi, 8 + \pi \right)$

Thus, the range of $f(x)$ is $\left( 8 - \pi, 8 + \pi \right)$.

Let me know if you need any clarifications or more details! Here are five related questions to expand your understanding:

1. How does changing the coefficient inside the arcsine or arctangent functions affect the domain or range?
2. What is the difference between the domain constraints of $\arcsin$ and $\arctan$?
3. How would the domain of $f(x)$ change if it involved an $\arccos$ function instead of an $\arcsin$?
4. What transformations generally affect the range of inverse trigonometric functions?
5. How does shifting a function vertically (like adding 7 or 8) affect the overall range?

Tip: Remember that the range of inverse trigonometric functions like arcsine and arctangent is always limited, and understanding their base ranges helps greatly with solving these problems!