determining which graphs are bitpartite/2-colorable and which are not I am having trouble understanding bipartite/$2$-colorable graphs. I was hoping someone can guide me through this question.

For the graphs given above, either prove that they are bipartite by showing that they are $2$-colorable, or prove that they are not $2$-colorable, and so they are not bipartite.

 A: A graph is 2-colorable if we can color each of its vertices with one of two colors, say red and blue, in such a way that no two red vertices are connected by an edge, and no two blue vertices are connected by an edge (a k-colorable graph is defined in a similar way). So to see if a graph is 2-colorable, the easiest way is to start by coloring a random vertex with blue. Then every vertex adjacent to it gets colored red. After that, every vertex adjacent to a red vertex gets colored blue, and so on and so forth. The case of 2-colorability is easier than say 3-colorability, because there are no choices to be made (at least in the case for connected graphs, which are the important cases). Once a single vertex is colored, the evolution of the coloring is determined.
So for example, if we colored the graph in a.), lets start by coloring vertex $a$ blue. The only vertex adjacent to $a$ is $e$, so then we color $e$ red. Finally, the remaining three vertices are adjacent to $e$, so they get colored blue, which gives us a valid 2-coloring of the graph. 
However, in the graph in e.), if we start by coloring $a$ blue, then both $b$ and $c$ have to get colored red. But these vertices are adjacent, so we will have two red vertices that are adjacent. We get the same problem if $a$ is colored blue, so we can conclude that this graph is not 2-colorable.
A: There is a highly useful theorem that you can use here:

Theorem: A (simple) graph is bipartite if and only if it contains no odd cycles.

Using this theorem, look to see if there are any odd cycles (triangles, hexagons, 7-gons, etc)  in any of your examples.  You can see a triangle in (c) and in (e), and you can see a hexagon in (g), so these are not bipartite.
Quick verification will show that there are no odd cycles in the others, so they must be bipartite.

Proof of the theorem:
$\Rightarrow)$ Suppose that $G$ is bipartite.  Then there exist partitions $A$ and $B$ such that $V(G) = A\cup B$ with $A\cap B=\emptyset$ and there are no edges with both ends in the same partition.
Suppose for a contradiction that there is an odd cycle, say $(v_1, v_2,\dots, v_k, v_1)$ with $k$ odd.  Without loss of generality, let $v_1\in A$.  It is clear then that for this cycle, $v_n\in A$ if $n$ odd and $v_n\in B$ if $n$ even.  But then since $k$ is odd, we have the edge $(v_k,v_1)$ which is an edge between two vertices both of which in $A$ contradicting our assumption that $G$ was bipartite.
$\Leftarrow)$  Let $G$ have no odd cycles.  Without loss of generality, suppose that $G$ is connected (else induct on the number of connected components).  Pick a vertex, $v$.  Then partition the vertices into two sets $A$ and $B$ where $A=\{x\in V(G)~:~\text{the shortest path from}~x~\text{to}~v~\text{is of even length}\}$ and $B=\{x\in V(G)~:~\text{the shortest path from}~x~\text{to}~v~\text{is of odd length}\}$.  Clearly then $A\cup B=V(G)$ and $A\cap B=\emptyset$ by construction.  We now wish to prove that there do not exist any edges between vertices in the same partition.
Suppose for a contradiction that there exists an edge between two vertices in the same partition, without loss of generality an edge $(a_1,a_2)$ with $a_1,a_2\in A$.  Then there is an odd length closed path $(v,\dots,a_1,a_2,\dots,v)$.  However, odd length closed paths must necessarily contain an odd cycle ($\star$), contradicting that $G$ contains no odd cycles.  $\square$

Proof of $(\star)$ left as an exercise to the reader.
A: I’ll take you through a couple of them to get you started.
The easy ones are the ones that are $2$-colorable: you just find a $2$-coloring, i.e., a coloring of the vertices using two colors in such a way that vertices of the same color are never adjacent along an edge. For example, (a) is $2$-colorable: if you color vertex $e$ red, you can color vertices $a,b,c$, and $d$ blue, and no two vertices of the same color will be adjacent. You shouldn’t have any trouble finding $2$-colorings of (f) and (h), and there are two more that are $2$-colorable as well.
Now look at (e): each vertex is adjacent to both of the others. Color $a$ red, and you have to color both $b$ and $c$ blue, but then the two blue vertices are adjacent. Can you see that you’ll run into this problem whenever your graph contains a triangle? That will take care of one other graph in the set.
If you get all that, only one graph will be left, and an argument similar to the one for the triangle will take care of it.
