The linear program you give as the dual is correct. However, the optimal solution isn't $g=0$, but rather $g=-6$ at $(w_1,w_2)=\left(0,-\frac{3}{5}\right)$.
Notice that $g=0$ isn't a possibility because if $g=0$ then we have $w_1=w_2=0$ which then does not satisfy the constraint
$$6w_1+5w_2\le-3$$
You can also notice that this is the only nontrivial constraint in the dual program - the other constraints are satisfied merely by the $w_1,w_2\le 0$ requirement.
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
By evaluating the objective function at each BFS of the primal, we see that $(x_1^\star,x_2^\star)=\left(\frac85,-\frac{24}5\right)$ is an optimal solution, with objective value $-8$. Consider the dual problem: \begin{align} \max & \quad 16y_1 +4y_2-24y_3\\ \mathrm{s.t.} &\quad 2y_1-\frac12 y_2-3y_3\geqslant -1\\ &\quad 4y_1 -y_2 +4y_3\leqslant \frac43\\ &\quad y_1,y_2,y_3\geqslant 0 \end{align} At optimality, $x_1$ and $x_2$ are nonzero, so by complementary slackness, the corresponding constraints in the dual are binding. Also, the constraint in the primal corresponding to the variable $y_1$ in the dual is not binding, so $y_1=0$ at optimality. It follows that $(y_2, y_3)$ is an optimal basis for the dual, with $y_2^\star = 0$, $y_3^\star = \frac13$, and objective value $8$. This is a feasible solution for the dual with the same objective value as an optimal solution for the primal, and therefore is an optimal solution for the dual.