For an equation of the form $|A|=8$, the expression inside the absolute value can be either $8$ or $-8$. So solve the two equations $2x-6=8$ and $2x-6=-8$. From $2x-6=8$, add 6 to get $2x=14$, so $x=7$. From $2x-6=-8$, add 6 to get $2x=-2$, so $x=-1$. Therefore, the equivalent expression for the solution set is $x=-1$ or $x=7$.

An absolute value graph in vertex form is $y=a|x-h|+k$, where the vertex is $(h,k)$. Since the vertex is $(3,-2)$, the equation must have the form $y=a|x-3|-2$. That narrows the choices to A. To confirm, use the point $(5,4)$:

So $2a=6$, which gives $a=3$. Therefore the equation is $y=3|x-3|-2$.

From $x-y=2$, write $x=y+2$. Substitute into $|x+y|=8$:

This gives two cases:

Case 1: $2y=6$, so $y=3$ and then $x=5$. Thus, $xy=5\cdot 3=15$.
Case 2: $2y=-10$, so $y=-5$ and then $x=-3$. Thus, $xy=(-3)(-5)=15$.
In both cases, the product is $15$, so the correct answer is $15$.

At $x=6$, the deviation is $y=|6-3|+1=3+1=4$. To find another time with the same deviation, solve $|x-3|+1=4$. Subtract 1 from both sides: $|x-3|=3$. This gives two solutions: $x-3=3$ or $x-3=-3$, so $x=6$ or $x=0$. Since the question asks for an additional time value, the answer is $0$.

The equation $|x|+|y|=10$ represents a square centered at the origin, but its sides are not parallel to the axes; instead, its vertices are on the axes at $(10,0)$, $(-10,0)$, $(0,10)$, and $(0,-10)$. To find where the line $y=3$ meets the boundary, substitute $y=3$ into the equation:

So the two intersection points are $x=7$ and $x=-7$, giving points $(7,3)$ and $(-7,3)$. The walkway runs horizontally between these two points, so its length is the difference in the $x$-coordinates:

Therefore, the length of the walkway is $14$.