Show that the following family of hash functions is $2$-wise independent but not $3$-wise independent
Hint for showing 2-wise independence: Given $x_1,x_2$ and $v_1,v_2$, write the desired equations in a different way:
$$
b = v_1 - Ax_1, \\
A(x_2-x_1) = v_2 - v_1.
$$

Regarding 3-wise independence: I could be missing something, but it seems that the family is 3-wise independent.

Hint for refuting 4-wise independence: Take $x_3 = x_1 + x_2$ and $x_4 = 0$.