Answer
1. $m\ne0$
2. $f^{-1}(x)=\frac{x-b}{m}$
3. Yes, $\frac{1}{m}$
Work Step by Step
1. A one-to-one function will intersect only once with any horizontal line. For the linear function
$f(x)=mx+b$, it requires that $m\ne0$ which means that it can not be a horizontal line itself.
2. Assume $m\ne0$ and $y=mx+b$, we can obtain $x=\frac{y-b}{m}$, switch $x,y$ to get the inverse as
$y=f^{-1}(x)=\frac{x-b}{m}$
3. It can be seen that the inverse is linear with a slope of $\frac{1}{m}$
