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.
There is such aspect.
Consider system of equations
$$
\begin{cases}
(1):x_1+4x_2 \le 3 \\
(2):x_2+4x_3 \le 2 \\
(3):x_1+2x_2+3x_3=5 \\
(4):x_2+x_3=1
\end{cases}
$$
Make some substitutions:
$$
\begin{align}
(4)\to (2)&: x_3\le\frac{1}{3}\\
(4)\to (3)&: x_1=3-x_3 \quad(5)\\
(4)&: x_2=1-x_3 \quad(6)\\
(5),(6)\to(1)&: x_3\ge\frac{4}{5}
\end{align}
$$
Notice that $x_3\le\frac{1}{3}$ and $x_3\ge\frac{4}{5}$ contradict each other, hence system is incompatible.
Best Answer
Assume $3x_1+2x_2+x_3=6$, plug that into the first inequality to derive $2x_2\leq 0$ implying $x_2=0$ using the first equation $$3x_1+x_3=6$$ we derive at $$x_3=6-3x_1$$
now intersect this with the feasible region (second inequality) and you should be done by using the boundary values of the set of solutions (note that these values are $\subset \mathbb{R}^3$) because the relationship is linear