Such a number should be in the form of $2^m3^n$ where $m$ and $n$ are non-negative integers such that $m$ and $n$ are not both zero.
If $n=0$, $m\ge1$ and $2^m\le1000$. So $1\le m\le 9$.
If $n=1$, $2^m\le\frac{1000}{3}$. So $0\le m\le 8$.
If $n=2$, $2^m\le\frac{1000}{9}$. So $0\le m\le 6$.
If $n=3$, $2^m\le\frac{1000}{27}$. So $0\le m\le 5$.
If $n=4$, $2^m\le\frac{1000}{81}$. So $0\le m\le 3$.
If $n=5$, $2^m\le\frac{1000}{243}$. So $0\le m\le 2$.
If $n=6$, $2^m\le\frac{1000}{729}$. So $m= 0$.
If $n=7$, $2^m\le\frac{1000}{2187}$, which is impossible.
Number of possibilities is $9+9+7+6+4+3+1=39$.
You've got some issues with your approach. Let me address them point by point.
Let $A = 3|n$ , $B = 4|n $ and $C = 3 \bullet 4 | n$ where $|C| = |A \cup B| = \lfloor\frac{|A|}{4}\rfloor = \lfloor\frac{|B|}{3}\rfloor$
One issue here is a bit of a nitpick: you seem to want $A,B,C$ to be statements and sets at the same time. Rather, I think you mean $A=\{n\in U:3\mid n\},B=\{n\in U:4\mid n\},C=\{n\in U:12\mid n\},$ where $U=\{n\in\Bbb Z^+:100\le n\le 999\}.$ Another issue may simply be a typo: we should have $C=A\cap B,$ not $C=A\cup B.$
I understood the formula as follows: $\left|A\cup \overline B\right| = |A| - |A \cup B| $ but can I transform this into the format: $|A\cup \overline B| = |A| + |\overline B| - |A \cap \overline B| $ or vice versa?
Here again, I suspect a typo: we should have $\left|A\cap \overline B\right| = |A| - |A \cap B|,$ assuming that $\overline B$ refers to the relative complement of $B$ in $U.$ Your second formula here is correct, but we aren't interested in the numbers that are divisible by $3$ or not divisible by $4,$ so it isn't relevant.
I tried the following to show show that $$\begin{eqnarray}|A \cup B| &=& |A| + |B| - |A \cap B|\\ &=& \left|A \cap \overline B\right| - \left|\overline B\right|\\ &=& \left|\overline{\overline{A \cap \overline B}}\right| - \left|\overline B\right|\\ &=& \left|\overline{\overline A \cup B}\right| - \left|\overline B\right|\\ &=& |U| - \left|\overline A \cup B\right| - \left|\overline B\right|\\ &=& |U| - \left|\overline A\right| - |B| - \left|\overline A \cap B\right| - \left|\overline B\right|...\end{eqnarray}$$ which I can't seem to finish.
This again doesn't seem to be relevant, but I'll talk about it, anyway. It is true that $|A \cup B| = |A| + |B| - |A \cap B|,$ since we're dealing with finite sets. By the formula $|A\cap \overline B| = |A| - |A \cap B|,$ we should then conclude that $$|A\cup B| = |A\cap \overline B|+|B|,$$ instead.
You've correctly applied DeMorgan's laws to $A\cap\overline B$, so that $$\left|\overline{\overline A \cup B}\right| - \left|\overline B\right| = |U| - \left|\overline A \cup B\right| - \left|\overline B\right|,$$ but then went off the rails again. Since $\left|\overline A \cup B\right|=\left|\overline A\right|+|B|-\left|\overline A\cap B\right|,$ we should instead find that $$|U| - \left|\overline A \cup B\right| - \left|\overline B\right|=|U|-\left|\overline A\right|-|B|+\left|\overline A\cap B\right|-\left|\overline B\right|.$$
Now, instead, we'll use the formula $$|A\cap \overline B| = |A| - |A \cap B|,$$ or more simply, $$|A\cap \overline B| = |A| - |C|.\tag{1}$$ All that's left is to calculate $|A|$ and $|C|.$ I'll take care of $A,$ and I'll leave $C$ to you.
Just to spoil it, I'll let you know that $$|A|=\frac{|U|}3=\frac{900}3=300.$$ But why is that? Note that in each of the sets $$\{100,101,102\},\{103,104,105\},\{106,107,108\},$$ there is exactly one element that is divisible by $3,$ and these sets have no elements in common More generally, for any integer $n,$ the set $\{99+3n-2,99+3n-1,99+3n\}$ has exactly one element divisible by $3,$ and if $m$ is an integer with $m\ne n,$ then $$\{99+3n-2,99+3n-1,99+3n\}\cap\{99+3m-2,99+3m-1,99+3m\}=\emptyset.$$ Now, note that $$U=\bigcup_{n=1}^{300}\{99+3n-2,99+3n-1,99+3n\},$$ so the number of elements in $A$ is the same as the number of $3$-element sets we've partitioned $U$ into, which is $\frac{900}3=300.$
Best Answer
$250$ numbers are divisible by $4$. Now use inclusion/exclusion principle:
$$250-\left\lfloor\frac{250}{3}\right\rfloor-\left\lfloor\frac{250}{4}\right\rfloor+\left\lfloor\frac{250}{3\times4}\right\rfloor=125$$
Please note that we use $4$ instead of $16$, because those $250$ numbers are not consecutive - they already contain only multiples of $4$, so we need to exclude only those that are multiples of yet "another $4$" (the credit for this fix goes to @AndréNicolas, as implied in the comment below).