Your formulation of the dual Problem is right-you forgot the negative sign at the third constraint only. As I said, you can eleminate the third constraint (primal problem).In this case the dual problem is:
$$
\text{min } 8y_1+10y_2\\
2y_1-y_2\ge 5\\
4y_1+y_2\ge 7\\
-2y_1+2y_2\ge -3\\
y_1,y_2\ge0
$$
The solution of this problem is $y^T=(y_1,y_2)=(3.5;2)$ Your solution is right. The solution is $y_3=0$, because the third constraint is not necessary.
The compementary slackness condition is $X^TC^*=b^TY^*$
This gives: $\begin{pmatrix}5 & 7 & -3\end{pmatrix}\cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 8 & 10 \end{pmatrix} \cdot \begin{pmatrix} 3.5 \\ 2 \end{pmatrix} \Rightarrow \begin{pmatrix}5 & 7 & -3 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=48$
Additional
The following condition must hold $x_j \cdot z_j=0 \ \ \forall n$
$z_j$ are the slack variables of the dual problem. If you insert the solution for the dual problem, then you will see, that $z_1$ and $z_3$ are zero and $z_2$ is not zero. Thus the equations are (for all n):
$x_1\cdot 0=0 $
$x_2\cdot z_2=0$
$x_3\cdot 0=0$
Thus you know for sure, that $x_2=0$
And secondly this equation must hold: $s_i\cdot y_i=0 \ \ \forall m$
$s_i$ are the slack variables of the primal problem. Thus $s_1$ and $s_2$ are zero and so we have the equations:
$2x_1-2x_3=8$
$-x_1+2x_3=10 $
Remember, that $x_2=0$
This two equations you can solve.
The solution of the primal is $x_{opt}^{T}=(4,3)$.
From the complementary slackness theorem we know:
$x_j\cdot z_j=0 \ \forall \ \ j=1,2, \ldots , n$
$y_i\cdot s_i=0 \ \forall \ \ j=1,2, \ldots , m$
$s_i$ are the slack variables of the primal problem.
$z_j$ are the slack variabales of the dual problem.
First constraint
$-2x_1-4x_2 +s_1= -12\Rightarrow -8-12+s_1=-12\Rightarrow s_1 >0$
Thus $y_1=0$
Fourth constraint
$x_2 \leq 5$
$3+s_4=5 \Rightarrow s_4=2>0$
Thus $y_4=0$
Both $x_i$ are greater than $0$. Thus $z_1=z_2=0$. We can take the first and second constraints and transform them into equations:
$-2y_1+y_2+y_3 = 6\\
-4y_1+y_2+y_4 = 4\\$
$0+y_2+y_3=6\\
0+y_2+0=4$
Now the solution is obvious.
Best Answer
Is $x=(1,3)$ feasible: yes, the constraints are satisfied.
Is $x=(1,3)$ optimal? Consider complementary slackness:
$\begin{align} y_1(2 x_1 + x_2 - 8) = 0 \\ y_2(x_1 + 2 x_2 - 7) = 0 \\ y_3(x_2 - 3) = 0 \\ x_1(2y_1 + y_2-4) = 0 \\ x_2(y_1 + 2y_2 + y_3-5) = 0 \end{align}$
The first two equalities yield $y_1=y_2=0$, but this violates the fourth equality. Since no $y \geq 0$ can be found that satisfies complementary slackness, $x$ is not optimal.