Edited hint: Since a reformulation of the original problem into standard form has now been added, my original answer is no longer very comprehensible. The dual of the problem, as originally formulated is:
$
\begin{eqnarray}
\max 5y_1\ + &y_2&+\ 2y_3\\
\ \ \ \ \ \ \ \ \mbox{ subject to:} & &\\
y_1+y_2\ \ \ &=& 2\\
y_1 + \ \ \ y_3&=&3\ \mbox{, and}\\
y_1\ge0, y_2&\ge0,& y_3\ge 0
\end{eqnarray}
$
As callculus has pointed out, in the absence of any explicit sign restrictions on the variabes $x_1, \mbox{and } x_2\ $, the constraints in the dual should be equations rather than inequalities. However, since the second and third constraints of the primal imply that $\ x_1\ge0, \mbox{and } x_2\ge0\ $, these implied constraints could have been made explicit in the original formulation—as you have now so made them in your reformulation into standard form—, and then the dual constraints would have been $y_1+y_2 \le 2\ $ and $y_1 + y_3\le3\ $, as they are, in fact, for the standard form, whose dual is:
$
\begin{eqnarray}
\max 5y_1\ + &y_2&+\ 2y_3\\
\ \ \ \ \ \mbox{ subject to:}& &\\
y_1+y_2\ \ \ &\le& 2\\
y_1 + \ \ \ y_3&\le&3\ \mbox{, and}\\
y_1\ge0, y_2&\ge0,& y_3\ge 0\ \ .
\end{eqnarray}
$
Now if $\ x^* = \left(3, 2\right)\ $ is an optimal solution for the primal, then because the non-negativity conditions, $\ x_1 \ge 0\ \mbox{ and } x_2 \ge0$, of your primal (in standard form) are both slack for those values (i.e. $\ 3\, {\color{red} >}\, 0\ \mbox{ and }\ 2\, {\color{red} >}\, 0 $), the complementary slackness conditions imply that the corresponding constraints, $\ y_1+y_2\le 2\ \mbox{ and } y_1 + y_3\le3\ $, must be tight for an optimal solution, $\ \left(y_1^*, y_2^*, y_3^*\right)\ $, of the dual—that is, $\ y_1^*+y_2^*= 2\ \mbox{ and } y_1^* + y_3^*=3\ $. Similarly, because the non-negativity condition $x_4 \ge 0$ is also slack (since $\ x_4 = x_1^* - 1 = 2\, {\color{red} >}\, 0\ $), then the corresponding constraint, $\ y_2 \ge 0\ $, should be tight for the optimal solution of the dual. That is, $\ y_2^* = 0\ $. These conditions are sufficient to determine uniquely what the optimal solution of the dual would have to be—namely, $\ y_1^* = 2, y_2^* = 0,\ $ and $\ y_3^* = 1\ $. To verify that these purported optimal solutions really are optimal, all you have to do is check that the values of their respective objective functions are the same for both of them, which is, in fact, the case.
