Questions concerning smallest fraction between two given fractions. I recently encountered this on a practice test

Find the smallest positive integer $n$ such that there exists an integer $m$ satisfying $0.33 < \frac{m}{n} < \frac{1}{3}$

My answer:

$$\begin{align}0.33 < \frac{m}{n} < \frac{1}{3}&\implies(\frac{33}{100} < \frac{m}{n})\land(\frac{m}{n}<\frac{1}{3}) \\ &\implies(33n<100m)\land(3m<n) \\\end{align}$$ and thus $n=3m+1$.
So $$\begin{align}33n<100m&\implies33(3m+1)<100m \\ &\implies 99m+33<100m\\ &\implies m>33\end{align}$$
Taking $m\geq34$, we find that $n=34\times3+1=103$

After the test, I had the following thoughts, which I do not know how to answer.
Questions
Will the solution for $n$ always yield the smallest $m$ possible?
Does this method (find minimum value of $n$ and substitute) work for all such problems? (where $m,n\in\mathbb{Z}$)
Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}<\frac{m}{n}<\frac{c}{d}$).

note: sorry for asking multiple questions in one post (which I know some people frown on) but I feel posting multiple questions with the same 'introduction' (problem+proof) would clutter and be somewhat cumbersome.
 A: 
Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}<\frac{m}{n}<\frac{c}{d}$).  

This answer uses your method.
In the following, $a,b,c,d,m,n$ are positive integers.
$\frac ab\lt \frac mn\lt \frac cd$ is equivalent to
$$na\lt mb\qquad\text{and}\qquad \frac{md}{c}\lt n$$
From the second inequality, we can set $n=\lfloor\frac{md}{c}\rfloor+1$ where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.
So, from the first inequality, we get
$$\left(\left\lfloor\frac{md}{c}\right\rfloor+1\right)a\lt mb\tag1$$
As a result, we can say that the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\frac ab\lt\frac mn\lt\frac cd$ is given by$$\left\lfloor\frac{Md}{c}\right\rfloor+1$$
where $M$ is the smallest integer $m$ satisfying $(1)$.

If $c=1$, then the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\frac ab\lt\frac mn\lt\frac 1d$ is given by$$\left(\left\lfloor\frac{a}{b-da}\right\rfloor+1\right)d+1$$
A: Yes to both of your questions. This is essentially the mediant for two terms in a Farey sequence. There is some deep theory here. Read up on Farey sequences for more information.
