Find the equations of the medians BQ and AM of triangle ABC, given A(-7, 8), B(-2, -2), and C(5, -1).

The coordinates of point $B$ are given from the previous problem.
The coordinates of B are $\boxed{(-2; -2)}$.

$Q$ is the midpoint of $AC$. The coordinates of $A$ are $(-7; 8)$ and $C$ are $(5; -1)$.
The midpoint formula is $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

Now, find the equation of the line passing through $B(-2; -2)$ and $Q(-1; \frac{7}{2})$.
First, calculate the gradient $m_{BQ}$:

Using the point-slope form $y - y_1 = m(x - x_1)$ with point $B(-2; -2)$:

Multiply by 2 to clear the fraction:

The equation of BQ is $\boxed{11x - 2y + 18 = 0}$.

Determine the equation of AM.
$M$ is the midpoint of $BC$. The coordinates of $B$ are $(-2; -2)$ and $C$ are $(5; -1)$.

Now, find the equation of the line passing through $A(-7; 8)$ and $M(\frac{3}{2}; -\frac{3}{2})$.
First, calculate the gradient $m_{AM}$:

Using the point-slope form $y - y_1 = m(x - x_1)$ with point $A(-7; 8)$:

Multiply by 17:

The equation of AM is $\boxed{19x + 17y - 3 = 0}$.

The centroid $G$ of a triangle with vertices $(x_A, y_A)$, $(x_B, y_B)$, and $(x_C, y_C)$ is given by the formula:

Using $A(-7; 8)$, $B(-2; -2)$, and $C(5; -1)$:

The coordinates of the centroid are $\boxed{\left(-\frac{4}{3}; \frac{5}{3}\right)}$.