Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The electric field is a vector quantity and we have here two point charges.
The electric field from $q_1$ is toward the charge while the electric field from $q_2$ is a way from it.
So at point to the left from $q_1$, we can find a point at which the electric field is zero. Also at a point to the right from $q_2$, we can find a point at which the electric field is zero as well.
But between the two charges the net electric field must be toward the left.
The net electric field to the left from $q_1$,
$$E_{net}=\dfrac{k|q_1|}{r_1^2}-\dfrac{k|q_2|}{r_2^2}=0$$
where $r_2=r_1+0.02$ where $0.02$ m is the distance between the two point charges.
$$ \dfrac{ \color{red}{\bf\not} k \color{red}{\bf\not} |q_1|}{r_1^2}=\dfrac{ \color{red}{\bf\not} k \color{red}{\bf\not} |q_2|}{(r_1+0.02)^2} $$
Hence,
$$(r_1+0.02)^2=r_1^2$$
$$r_1^2+0.04r_1+0.02^2=r_1^2$$
$$r_1=\dfrac{-0.02^2}{0.04}=-0.01\;\rm m=-1\;\rm cm$$
which is an imaginary root result, so it is dismissed.
By the same approach, for a point to the right from $q_2$,
$$E_{net}=-\dfrac{k|q_1|}{r_1^2}+\dfrac{k|q_2|}{r_2^2}=0$$
where $r_1=0.02+r_2$
$$ \dfrac{ \color{red}{\bf\not} k \color{red}{\bf\not} |q_1|}{(r_2+0.02)^2}=\dfrac{ \color{red}{\bf\not} k \color{red}{\bf\not} |q_2|}{r_2^2} $$
Hence, $$r_2=\dfrac{-0.02^2}{0.04}=-0.01\;\rm m=-1\;\rm cm$$
which is an imaginary root result as well, so it is dismissed.
Now it is obvious from the mathematical manipulation that the net electric field will never be zero except at $\infty$. See the figures below.
$$\color{blue}{\bf [b]}$$
The electric potential is not a vector quantity, it is a scalar quantity.
$$V_{net}=V_1+V_2=0$$
Hence,
$$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$$
Since it is a scalar quantity, and since $q_1$ is negative and $q_2$ is positive, the net electric potential will not be zero except between the two charges and just in the middle. You can test that mathematically on your own. So, $r_1=r_2$
This means that, $V_{net}=0$ at $x=0$ and at $x=\pm \infty$
when the point is to the right from $q_2$
$$\dfrac{kq_1}{r_1}=\dfrac{-kq_2}{r_2} $$
where $r_1= r_2+0.02$;
$$\dfrac{ q_1}{r_2+0.02}=\dfrac{- q_2}{r_2} $$
$$\dfrac{ -2\times 10^{-9}}{r_2+0.02}=\dfrac{- 2\times 10^{-9}}{r_2} $$
Hence,
$$r_2=r_2+0.02$$ which is only be satisfied at $x=\pm \infty$
$$\color{blue}{\bf [c]}$$
See the figures below.
