One can use the geometry. However, the point of the sketch below is to show that the problem can be solved mechanically.
We could let the sides of the regular hexagon be, say, $a$. However, it is easier, at least for typing, to choose a specific length. Then we can if necessary scale our answer. Sides $1$ are not a bad idea, but sides $4$ are more convenient, fewer fractions for a while.
Let the $x$-axis be along $AB$, wjth $A=(-2,0)$ and $B=(2,0)$. The coordinates of $C$ are now easy to find. They are $(2+4\cos(60^\circ), 4\sin(60^\circ))$, that is, $(4, 2\sqrt{3})$.
Now that we know the coordinates of $A$ and $C$, we can find the equation of line $AC$.
Do a similar calculation for the other side. The coordinates of $D$ are, by symmetry, $(-4, 2\sqrt{3})$. To get to $G$, we add $2\cos(60^\circ)$ to the $x$-coordinate, and $2\sin(60^\circ)$ to the $y$-coordinate. Thus $G$ has coordinates $(-3,3\sqrt{3})$.
Now that we know the coordinates of $B$ and $G$, we can find the equation of line $BG$.
To find the coordinates of $S$, find where our two lines with known equations meet.
Now that we know the coordinates of $S$, we can find answers to the questions. The only one that is clear is the asked ratio between the area of $\triangle ABS$ and the area of the hexagon. Each area can be computed. For example, the area of $\triangle ABS$ is one-half of $4$ times the $y$-coordinate of $S$.