Why is $R_2$ and $ R_3$ parallel with $R_1$ in this circuit? 
I know that if $V_2$ wasn't there it would make sense if $R_2$ and $R_3$ were parallel with each other and in series with $R_1$. Why is it different in this case?
 A: The question was not very clear. 
If $V_2$ was not there, you simply combine $R_2$ and $R_3$ to $R_x$ as $1/R_x=1/R_2+1/R_3$ and sum $R_x$ with $R_1$ and you get simply $I_1$. Inserting $V_2$ creates another current competing with $I_1$.
Concerning the boxed statement, I believe that it reflects the sums of voltages, not really it combines the resistors, seems ok to me.
A: The law of currents says that the sum of currents in any node/junction is zero. The law of voltages says that the sum of emf=sum(i*R) around any closed loop and equals the sum of voltage sources; sum(e) in the loop. According to this your equations on the right of the figure are correct. The rest is simple mathematical manipulations. For complex circuits all you need is repeat the same for every node and every loop and get the number of equations equal to the number of unknowns and solve. 
A: Always remember that you are free to deform and stretch a circuit without changing its topological properties, i.e. no deleting or adding nodes, or popping them over circuit elements.
The easiest way to see that R1, R2, and R3 are all in parallel is to pull R1 to the left along its wire until it is vertical, and similarly pull R2 to the right along its wire until it is vertical. Then the elements should be parallel by visual inspection.
Regarding the presence or absence of V2 - this has absolutely nothing to do with it. The relationship of one circuit to an element to another (series, parallel, or some combination of the two) is entirely determined by the the two of them and the wires connecting them; no other circuit element changes this topological relationship.
It's sometimes very hard to tell by eye if circuit elements are parallel to each other or in series. There might be algorithmic ways of doing so, as in the other answers, but I prefer to use topological freedom to redraw circuits in a standard way where it is easier for me to tell parallel and series relationships apart.
A: They aren't.  Two resistors are in parallel if they have the same voltage drop across them; two resistors are in series if the same current flows through them.  
In your problem, because of the presence of $v_2$, it's entirely possible to have different currents in and voltage drops across all of $R_1,R_2,R_3$.
I believe your boxed statement,
\begin{align}
v_1 &= i_i (R_1 + R_3) - i_2 R_3 \\
-v_2 &= -i_1 R_3 + i_2 (R_2 + R_3)
\end{align}
is correct, but notice that each equation has $R_3$ twice.  If anything that suggests $R_3$ is in parallel with itself, which is of course not physical.
A: in the absence of the voltage V2, R2 and R3 will be parallel so R(equivalent)=(R3*R2)/(R3+R2) ,and R(equivalent) which im gonna call R' is in serie with R1 so you circuit will contain just the voltage V1 and a resistor R''=R'+R1.
in the presence of the voltage V2, R1,R2 and R3 are in wye "Y" ,get a look to this http://en.wikipedia.org/wiki/Y-%CE%94_transform
