Questions 7, 8 and 9 are about algebra, probability and geometry respectively.

Given the relation:

$$y = \frac{4m(k+x)}{k+2x}$$

Multiply both sides by $(k+2x)$:

$$y(k+2x) = 4m(k+x)$$

Expand both sides:

$$yk + 2xy = 4mk + 4mx$$

Gather terms containing $k$ on one side and terms without $k$ on the other side:

$$2xy - 4mx = 4mk - yk$$

Factor out $k$ from the right side:

$$2xy - 4mx = k(4m - y)$$

Divide by $(4m - y)$ to isolate $k$:

$$k = \frac{2xy - 4mx}{4m - y}$$

So, the subject of the relation is $\boxed{k = \frac{2xy - 4mx}{4m - y}}$.

Substitute the given values $x=3$, $y=-2$, and $m=1$ into the expression for $k$:

$$k = \frac{2(3)(-2) - 4(1)(3)}{4(1) - (-2)}$$

Calculate the numerator:

$$2(3)(-2) - 4(1)(3) = -12 - 12 = -24$$

Calculate the denominator:

$$4(1) - (-2) = 4 + 2 = 6$$

Now, divide the numerator by the denominator:

$$k = \frac{-24}{6}$$ $$k = -4$$

The value of $k$ is $\boxed{-4}$.