Stalk of the Restricted Sheaf Given a sheaf $F$ on a topological space $X$ and $U$ is an open subset of $X$. Denote $F|_U$ be the restricted sheaf of $F$. Then to any $y\in U$. Do we have $F_y=(F|_U)_y$?
 A: @Bueggi has the right answer from first principles. I'd like to mention a slightly more abstract way to think about it, which I hope will be helpful to see if you're starting to learn about sheaves.
If $*$ is a topological space with one point, the category sheaves on $*$ is canonically isomorphic to the category of sets by taking global sections, that is, $F \mapsto F(*) : \mathsf{Sh}(*) \xrightarrow{\Gamma} \mathsf{Set}$ is an isomorphism of categories.
Now, if $X$ is a topological space and $p \in X$, there is a unique continuous map $i_p : * \to X$ whose image is $\{p\}$. For any sheaf $F$ on $X$, $F_x$ is precisely the set of global sections of the inverse image sheaf $i_p^{-1} F$ — you should try to prove this! In other words, the composition $\mathsf{Sh}(X) \xrightarrow{i_p^{-1}} \mathsf{Sh}(*) \xrightarrow{\Gamma} \mathsf{Set}$ is equal to the stalk functor $F \mapsto F_p : \mathsf{Sh}(X) \to \mathsf{Set}$ (this requires also checking what happens to morphisms).
From this perspective, it's easy to see what happens in your question: let $p \in U$ where $U$ is an open subset of $X$. Then the map $i_p : * \to X$ factors as $i_U \circ j_p$ where $j_p : * \to U$ has image $\{p\}$ and $i_U : U \to X$ is the inclusion. By functoriality of the inverse image construction, we have that $i_p^{-1} = (i_U \circ j_p)^{-1} = j_p^{-1} \circ i_U^{-1}$. If $F$ is a sheaf on $X$, then $i_U^{-1}(F) = F|_U$, so $i_p^{-1}(F) = j_p^{-1}(F|_U)$. The set of global sections of the left-hand side is just $F_p$, and the set of global sections of the right-hand side is $(F|_U)_p$.
To rephrase: taking the stalk at $p$ is equivalent to pulling back to the point $\{p\}$. Pulling back sheaves is a functorial construction, so it doesn't matter if pull back directly to $\{p\}$ or pull back to $U$ first and then pull back to $\{p\}$. The pullback of a sheaf to an open subset is just the restriction, so we have the desired result.
This point of view can be helpful for thinking about more than just stalks of restrictions; for example this argument shows that pulling back a sheaf "preserves its stalks" in the following sense:
Fact Let $f : X \to Y$ be a continuous map between topological spaces. Let $F$ be a sheaf on $Y$ and let $x \in X$. Then $(f^{-1}F)_x = F_{f(x)}$.
A: Yes we have. 
To sections define the same germ  if they agree on  an open neighborhood of a point p. So they will also agree on  a smaller neighborhood.
