Let's begin with the functions:

$f(x) = x - 8 \quad \text{and} \quad g(x) = 9x^2$

We'll go through parts (a)-(h) step by step:

Part (a): $(f+g)(x)$

The sum of two functions is defined as: $(f+g)(x) = f(x) + g(x)$ Substitute the given functions: $(f+g)(x) = (x - 8) + 9x^2 = 9x^2 + x - 8$

Domain of $(f+g)(x)$:

Since both functions $f(x)$ and $g(x)$ are defined for all real numbers, their sum is also defined for all real numbers. So, the domain of $(f+g)(x)$ is: $\text{Domain: } (-\infty, \infty)$

Part (b): $(f-g)(x)$

The difference of two functions is defined as: $(f-g)(x) = f(x) - g(x)$ Substitute the given functions: $(f-g)(x) = (x - 8) - 9x^2 = -9x^2 + x - 8$

Domain of $(f-g)(x)$:

Since both functions are defined for all real numbers, their difference is also defined for all real numbers. So, the domain of $(f-g)(x)$ is: $\text{Domain: } (-\infty, \infty)$

Part (c): $(fg)(x)$

The product of two functions is defined as: $(fg)(x) = f(x) \cdot g(x)$ Substitute the given functions: $(fg)(x) = (x - 8)(9x^2) = 9x^3 - 72x^2$

Domain of $(fg)(x)$:

The product is defined for all real numbers. So, the domain of $(fg)(x)$ is: $\text{Domain: } (-\infty, \infty)$

Part (d): $\left(\frac{f}{g}\right)(x)$

The quotient of two functions is defined as: $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$ Substitute the given functions: $\left(\frac{f}{g}\right)(x) = \frac{x - 8}{9x^2}$

Domain of $\left(\frac{f}{g}\right)(x)$:

For this quotient to be defined, $g(x) \neq 0$. Set $g(x) = 9x^2 = 0$, which gives $x = 0$. Therefore, the function is undefined at $x = 0$.

So, the domain of $\left(\frac{f}{g}\right)(x)$ is: $\text{Domain: } (-\infty, 0) \cup (0, \infty)$

Part (e): $f(3)$

Substitute $x = 3$ into $f(x)$: $f(3) = 3 - 8 = -5$

Part (f): $g(3)$

Substitute $x = 3$ into $g(x)$: $g(3) = 9(3^2) = 9(9) = 81$

Part (g): $(f+g)(3)$

Use the expression for $(f+g)(x)$ and substitute $x = 3$: $(f+g)(3) = 9(3^2) + 3 - 8 = 81 + 3 - 8 = 76$

Part (h): $\left(\frac{f}{g}\right)(3)$

Use the expression for $\left(\frac{f}{g}\right)(x)$ and substitute $x = 3$: $\left(\frac{f}{g}\right)(3) = \frac{3 - 8}{9(3^2)} = \frac{-5}{81}$

$\boxed{\frac{-5}{81}}$

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Let me know if you'd like more details or explanations. Here are some related questions for further exploration:

1. What is the inverse of $f(x) = x - 8$?
2. What is the range of $g(x) = 9x^2$?
3. How would you find the intersection points of $f(x)$ and $g(x)$?
4. What is the derivative of $(f+g)(x)$?
5. How do you find the domain of more complex composite functions like $f(g(x))$?

Tip: Always check for undefined points (such as where the denominator equals zero) when working with quotients of functions.