Pullbacks along monomorphism are pushouts in the category of graph and morphisms? In the category of graphs and morphisms, if the following diagram is a pullback diagram, is it also a pushout diagram? and vice versa? if so, how can I prove that? 

In the above diagram, $i$ and $i'$ are monomorphisms (inclusion morphisms)
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The following explanation is sketching of the graph that is provided as an example in the @Nex answer:
As seen in the following picture $C$ and $V'$ are equal to a graph with two vertices and one edge, and V just includes two vertices; however, the pushout of this diagram does not seem to be the same $V'$; the diagram pushout seems to be a graph that have two edges (instead of one) as the following:

 A: Edit The answer below assumed that the question was about graphs in the sense of https://en.wikipedia.org/wiki/Graph_(discrete_mathematics)#Graph rather than in the sense of what might be called a directed multigraph. However the first part is still valid in either case. In addition a final part is added below to address the second question for directed multigraphs.
The answer to your first question is no.
If $C=V=V'$ is the empty graph (the initial object in this category), then the diagram is a pullback (the only morphism to the empty graph is from the empty graph). However it will only be  a pushout if $C'$ is also the empty graph (since the universal property reduces to showing for each graph $G$ there should be a unique morphism $C'\to G$). 
The answer to your second question is no.
If $C=V'=C'$ is the graph with two vertices and one edge, $i=b'$ is the identity morphism, $V$ is the graph with two vertices and no edges, and $b=i'$ is the inclusion of $V$ in $C=V'$, then the diagram is a pushout (since any two graph morphism $f,g: C \to G$ which are equal after composing with the inclusion $b=i'$ must be equal). However it is not a pullback (since the pullback of an isomorphism is an isomorphism).
Edit The answer to your second question for directed multigraphs (assuming we are allowed self edges) is yes.To see why note that the category of such graphs is essentially a functor category of set. Let $\mathbf{I}$ be the category with two objects $E$ and $V$ and two non-identity morphisms $d: E\to V$ and $c:E\to V$. Then a functor $F$  from $\mathbf{I}$ to the category of sets, is determined by two sets $F(E)$ (the edge set) and $F(V)$ (the vertex set), and by two maps $F(d)$ (the domain) and $F(c)$ (the codomain). Writing $\mathbf{Set}$ for the category of sets one easily checks that the category of graphs is isomorphic to $\mathbf{Set}^\mathbf{I}$. Therefore, since limits and colimits are computed component-wise in functor categories (provided the limits/colimits of the components exists) it follows that your question reduces to the same question about such diagrams of sets and maps. To prove the result in set you might find it useful to note $i'$ being a monomorphism is a coproduct inclusion.
