Given the details you've provided, your student number is 30, so the last digit is 0. You've also mentioned that $g(x) = 0^{-x}$. Let's proceed with solving the expressions and determining the domain for each.

a. $g(x) - m(x)$

1. Given functions:

   • $g(x) = 0^{-x}$
   • $m(x) = \sin x$ (since the last digit is 0, $m(x) = \sin x$)

2. Simplified form:

   • $g(x) - m(x) = 0^{-x} - \sin x$

3. Domain:

   • The function $g(x) = 0^{-x}$ is not valid because $0^{-x}$ (where $x \neq 0$) leads to division by zero, which is undefined.
   • So, the domain is undefined.

b. $f(x)p(x)$

1. Given functions:

   • $f(x) = \sqrt{x}$
   • $p(x) = \frac{1}{x-1}$

2. Simplified form:

   • $f(x) \cdot p(x) = \sqrt{x} \cdot \frac{1}{x-1} = \frac{\sqrt{x}}{x-1}$

3. Domain:

   • $\sqrt{x}$ is defined for $x \geq 0$.
   • $\frac{1}{x-1}$ is defined for $x \neq 1$.
   • Therefore, the domain is $x \geq 0$ and $x \neq 1$.

c. $\frac{h(x)}{p(x)}$

1. Given functions:

   • $h(x) = \log (1-x)$
   • $p(x) = \frac{1}{x-1}$

2. Simplified form:

   • $\frac{h(x)}{p(x)} = \frac{\log(1-x)}{\frac{1}{x-1}} = (x-1) \log(1-x)$

3. Domain:

   • $\log(1-x)$ is defined for $1-x > 0$ which implies $x < 1$.
   • $p(x) = \frac{1}{x-1}$ is defined for $x \neq 1$.
   • Therefore, the domain is $x < 1$.

Would you like any further details or have any questions?

Here are five related questions to explore further:

1. How does changing the last digit affect each expression and its domain?
2. What would happen if $g(x)$ were different, for instance, $g(x) = 2^{-x}$?
3. How do logarithmic and sine functions compare in their domains and ranges?
4. How can the undefined nature of $0^{-x}$ be explained in different contexts?
5. How can we further simplify or analyze the expression $(x-1) \log(1-x)$?

Tip: Always check for restrictions in the domain, particularly when dealing with logarithmic, square root, or rational functions to avoid undefined expressions.