$f$ and $g$ are holomorphic function , $A=\{z:{1\over 2}<\lvert z\rvert<1\}$, $D=\{z: \lvert z-2\rvert<1\}$ $f$ and $g$ are holomorphic function defined on $A\cup D$ 
$A=\{z:{1\over 2}<\lvert z\rvert<1\}$, $D=\{z:\lvert z-2\rvert <1\}$ 


*

*If $f(z)g(z)=0\forall z\in A\cup D $ then either $f(z)=0\forall z\in A$ or $g(z)=0\forall z\in A$

*If $f(z)g(z)=0\forall z\in D $ then either $f(z)=0\forall z\in D$ or 
$g(z)=0\forall z\in D$

*If $f(z)g(z)=0\forall z\in A\cup D $ then either $f(z)=0\forall z\in A\cup D$ or $g(z)=0\forall z\in A\cup D$

*If $f(z)g(z)=0\forall z\in A $ then either $f(z)=0\forall z\in A$ or 
$g(z)=0\forall z\in A$
Except The $1,3$, I can say $2,4$ are true beacuse of Identity Theorem right? zeroes of holomorphic function are isolated. for $3$ I can define $f(z)=z$ on $A$ and $0$ on $D$ and $g(z)=0$ on $A$ and $z^2$ on $D$

You can use the identity theorem, but since that theorem doesn't directly apply to products, you should explain how you use the identity theorem.
You can, for example, argue that if $f\cdot g = 0$ on some connected open set $S$, then there are $S_1,S_2$ with $f = 0$ on $S_1$, $g = 0$ on $S_2$, $S_1 \cap S_2 = \emptyset$ and $S_1 \cup S_2 = S$. Now if $f$ isn't identically zero on $S$, then due to the identity theorem $S_1$ contains only isolated points, hence for every $x \in S_1$ you have that $B_\epsilon(x) \setminus {x} \subset S_2$ for some $\epsilon > 0$. Using the identity theorem again, you get $g = 0$ on $S$. The same applies with $f$ and $g$ reversed, thus at least one of them is identically zero on $S$.
This works, as you correctly stated, for (2) and (4). As Robert Isreal commented, you get (1) from (4), and for (3) your counter-example indeed refutes the assertion.
