Totally busted euler trail. New to graph theory proofs would like some help.
Let $G_{1}$ and $G_{2}$ be Eulerian graphs . let $v_{1}$ be a vertex in $G_{1}$ and $v_{2}$ be a vertex in $G_{2}$  jion $v_{1}$ and $v_{2}$ by a single edge.
1)Does the graph have a Euler trail? 
EDIT:
I think i finally get this part start at $v_{1}$ make a Euler circiut back to $v_{1}$ ignoring the new path to $v_{2}$ now delete these edges and cross form $v_{1}$ to $v_{2}$ now complete that euler circiut jion the ends together of these 2. Now as shown this is actually a Eulerian Trail. Which is where my confusion arose form.
2) is the resulting graph eulerian? why or why not?
The requirements for a graph to be Eulerian is that every vertex is of even degree, and clearly this was true in $G_{1}$ and $G_{2}$ as they where Eulerian thus all vertex of $G_{1}$ and $G_{2}$ where even. even +1 is not even hence there The new Graph is not Eulerian. ($ v_{1}$ and $v_{2}$ must both be of odd degree)
 A: It is important that the definition of an Eulerian graph here includes the property "is connected".  Without this property, we can have the following situation where $G_1$ is the blue graph and $G_2$ is the red graph, and we add in the orange edge:

Both $G_1$ and $G_2$ contain an Eulerian cycle, but the resultant graph does not contain an Eulerian trail.
We conclude that we must use the assumption that $G_1$ and $G_2$ are connected, otherwise the claim is not true due to the above counterexample.

For 1., you seem to be confusing Eulerian trail with Eulerian circuit.  Assuming $G_1$ and $G_2$ are connected, there is an Eulerian trail (and even an Eulerian circuit) starting at any vertex, and we can join them together as described. 
You can also approach 1. theoretically by using the following result:

An undirected graph has an Eulerian trail if and only if at most two vertices have odd degree, and if all of its vertices with nonzero degree belong to a single connected component.  (Source: Wikipedia)

Since the vertices of $G_1$ and $G_2$ have even degrees, the resultant graph has two odd degree vertices.  It is also connected since $G_1$ and $G_2$ are connected.  So the resultant graph satisfies the above condition, and thus contains an Eulerian trial.
Your approach to 2. seems natural, but I don't think the conditions are right, e.g. this non-Eulerian graph has every vertex of even degree:

This is what it should be:

An undirected graph has an Eulerian cycle if and only if every vertex has even degree, and all of its vertices with nonzero degree belong to a single connected component.  (Source Wikipedia, op. cit.)

A: 1) Yes. Take the Eulerian circuit, and delete that edge. You get a eulerian path.
2) No. Use the classification that a graph is Eulerian if and only if every vertex has even degree.
