solving $0<1-(3x/4)<1$ for $x$ My professor wrote this:
$$0<1-\frac{3x}{4}<1$$
$$-1<-\frac{3x}{4}<0$$
$$4>3x>0$$
$$\frac{4}{3}>x>0$$
Is that correct?

Note: This refers to an earlier version of the question, which has now been substantially changed.
Your professor’s step from
$$0<\frac{1-3x}4<1$$
to $$-1<\frac{-3x}4<0\;,$$
is incorrect: $1$ has been subtracted from $0$ and from $1$, but only $\frac14$ has been subtracted from $\frac{1-3x}4$. Had your professor correctly subtracted $\frac14$ from each term, the result would have been $$-\frac14<-\frac{3x}4<\frac34\;.$$
The next step was to multiply everything by $-4$; doing this to the corrected version leaves you with $$1>3x>-3\;,$$ and dividing everything by $3$ then gives you the result, $$\frac13>x>-1\;.$$
The first step of your calculation is clearly wrong: you’ve multiplied the $1$ by $4$ but not the $\frac{1-3x}4$. Judging by your next step, however, I think that you may have meant to multiply everything by $4$ to get $0<1-3x<4$. Then you wanted to subtract $1$, which is fine, but you must subtract it from the $0$ as well as from the $1-3x$ and the $4$, so you should get $-1<-3x<3$. Dividing everything by $-3$ will now give you
$$\frac13>x>-1\;,$$
as before.
