Let's analyze and solve the given problems step by step.

Functions given:

• $f(x) = \frac{1}{x}$
• $g(x) = x - 7$

Part (a): Find the domain of the following functions:

1. Domain of $f(x) = \frac{1}{x}$:

   • $f(x)$ is undefined when $x = 0$, because division by zero is undefined.
   • Therefore, the domain of $f(x)$ is $x \in (-\infty, 0) \cup (0, \infty)$.

2. Domain of $g(x) = x - 7$:

   • $g(x)$ is defined for all real numbers, so the domain is $(-\infty, \infty)$.

3. Domain of $f + g$:

   • $f + g$ is the sum of $f(x)$ and $g(x)$, i.e., $f(x) + g(x) = \frac{1}{x} + (x - 7)$.
   • The domain of $f + g$ is restricted by the domain of $f(x)$, which excludes $x = 0$.
   • So, the domain of $f + g$ is $x \in (-\infty, 0) \cup (0, \infty)$.

4. Domain of $f - g$:

   • $f - g$ is the difference of $f(x)$ and $g(x)$, i.e., $f(x) - g(x) = \frac{1}{x} - (x - 7)$.
   • The domain is the same as for $f + g$, which is $x \in (-\infty, 0) \cup (0, \infty)$.

5. Domain of $f \cdot g$ (product $fg$):

   • $f(x) \cdot g(x) = \frac{1}{x} \cdot (x - 7) = \frac{x - 7}{x}$.
   • Again, this function is undefined at $x = 0$, so the domain is $x \in (-\infty, 0) \cup (0, \infty)$.

6. Domain of $f \circ f$ (composition $ff$):

   • $f(f(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x$.
   • Since $f(x)$ is only defined for $x \neq 0$, the domain of $f \circ f$ is $x \in (-\infty, 0) \cup (0, \infty)$.

7. Domain of $\frac{f}{g}$:

   • $\frac{f(x)}{g(x)} = \frac{\frac{1}{x}}{x - 7} = \frac{1}{x(x - 7)}$.
   • This function is undefined when $x = 0$ or $x = 7$ (since $g(x) = 0$ when $x = 7$).
   • Therefore, the domain is $x \in (-\infty, 0) \cup (0, 7) \cup (7, \infty)$.

8. Domain of $\frac{g}{f}$:

   • $\frac{g(x)}{f(x)} = \frac{x - 7}{\frac{1}{x}} = x(x - 7)$.
   • This function is undefined when $x = 0$.
   • Therefore, the domain is $x \in (-\infty, 0) \cup (0, \infty)$.

Part (b): Find the following:

1. $(f + g)(x)$:

   • $(f + g)(x) = \frac{1}{x} + (x - 7) = \frac{1 + x(x - 7)}{x} = \frac{1 + x^2 - 7x}{x} = \frac{x^2 - 7x + 1}{x}$.

2. $(f - g)(x)$:

   • $(f - g)(x) = \frac{1}{x} - (x - 7) = \frac{1 - x(x - 7)}{x} = \frac{1 - x^2 + 7x}{x} = \frac{-x^2 + 7x + 1}{x}$.

3. $(fg)(x)$:

   • $(fg)(x) = f(x) \cdot g(x) = \frac{1}{x} \cdot (x - 7) = \frac{x - 7}{x}$.

4. $(ff)(x)$:

   • $(ff)(x) = f(f(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x$.

5. $\left(\frac{f}{g}\right)(x)$:

   • $\left(\frac{f}{g}\right)(x) = \frac{\frac{1}{x}}{x - 7} = \frac{1}{x(x - 7)}$.

6. $\left(\frac{g}{f}\right)(x)$:

   • $\left(\frac{g}{f}\right)(x) = \frac{x - 7}{\frac{1}{x}} = x(x - 7) = x^2 - 7x$.

Let me know if you want further clarification on any of these steps!

Here are five related questions to extend the problem:

1. What is the range of $f(x) = \frac{1}{x}$ and $g(x) = x - 7$?
2. How would you find the inverse functions of $f(x)$ and $g(x)$?
3. What is the composition $g(f(x))$, and what is its domain?
4. What is the behavior of $f(x) + g(x)$ as $x \to \infty$?
5. How do you simplify $f(x) \cdot f(g(x))$?

Tip: Always check for restrictions in the domain when dealing with fractions and compositions of functions to avoid undefined expressions.