First find the slope using the two points:

So the line has slope $-2$. Use one of the points to find the $y$-intercept in the form $y=mx+b$. Substituting $(2,7)$ gives

so

and therefore

The equation of the line is

Because line $k$ intersects line $m$ at $(2,5)$, the point $(2,5)$ is on line $k$. The stem also says line $k$ passes through $(4,1)$. So the slope of line $k$ is

Using slope-intercept form $y=mx+b$ with slope $-2$ and point $(2,5)$,

so

and

Therefore, the equation of line $k$ is

Because $f$ is linear, its slope can be found from the table. From $x=-1$ to $x=2$, the change in $x$ is $3$ and the change in $f(x)$ is $1-7=-6$, so the slope is $\frac{-6}{3}=-2$. Since $g$ is parallel to $f$, $g$ also has slope $-2$. Using the point $(4,10)$ on $g$, write $g(x)=-2x+b$. Then $10=-2(4)+b$, so $10=-8+b$ and $b=18$. Therefore, $g(0)=18$.

For two parallel lines, the perpendicular distance between them can be found by writing both in standard form. The lines are

Rewrite them as

For parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$, the distance between them is

So the width is

Rationalizing the denominator gives

So the width of the garden is $\frac{8\sqrt{5}}{5}$.

Let the total cost be $y=mx+b$, where $m$ is the cost per class and $b$ is the membership fee. Using the points $(3,39)$ and $(7,67)$, the slope is

So each class costs $7$. Substitute into the equation using $(3,39)$:

Therefore,

The membership fee is \18$.$