Answer
In Pair (1) and Pair (2), there are infinitely many points off the axis at which $V_{net} = 0$
Work Step by Step
We can write a general equation for the net electric potential:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
Pair (1):
Let $d_1$ be the distance between a point and the $-2q$ charge.
Let $d_2$ be the distance between a point and the $+6q$ charge.
We can find $V_{net}$ at this point:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
$V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{-2q}{d_1}+\frac{6q}{d_2})$
If $~~\frac{-2q}{d_1}+\frac{6q}{d_2} = 0~~$ then $~~V_{net} = 0$
If $~~d_2 = 3d_1~~$ then $~~V_{net} = 0$
There are infinitely many points off the axis such that $~~d_2 = 3d_1~~$
Pair (2):
Let $d_1$ be the distance between a point and the $+3q$ charge.
Let $d_2$ be the distance between a point and the $-4q$ charge.
We can find $V_{net}$ at this point:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
$V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{+3q}{d_1}-\frac{4q}{d_2})$
If $~~\frac{+3q}{d_1}-\frac{4q}{d_2} = 0~~$ then $~~V_{net} = 0$
If $~~d_2 = \frac{4d_1}{3}~~$ then $~~V_{net} = 0$
There are infinitely many points off the axis such that $~~d_2 = \frac{4d_1}{3}$
In Pair (3), $V_{net} \gt 0$ at all points since both charges are positive.
In Pair (4), $V_{net} \lt 0$ at all points since both charges are negative.
Therefore:
In Pair (1) and Pair (2), there are infinitely many points off the axis at which $V_{net} = 0$
