Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
When the two currents are in the same direction, the magnetic field between the two wires could be zero at some point, but it will never be zero outside them.
Note that the direction of the magnetic fields of both wires to the right from the right wire is upward and to the left from the left wire is downward.
So it will never be zero at $x\gt 2$ cm and at $x\lt -2$ cm.
See the figure below.
$$\sum F_B=0=B_1-B_2$$
We know that the magnetic field of a long straight current wire is given by
$$B=\dfrac{\mu_0I}{2\pi d}$$
So,
$$B_1-B_2=\dfrac{\mu_0I_1}{2\pi d_1}-\dfrac{\mu_0I_2}{2\pi d_2} =0$$
Let's assume that $\sum B=0$ at point P that is at distance $x$ from the origin, as seen below. So that $d_1=0.02+x$, and $d_2=0.02-x$.
$$ \dfrac{\mu_0I_1}{2\pi ( 0.02+x)}-\dfrac{\mu_0I_2}{2\pi (0.02-x)} =0$$
Solving for $x$,
$$ \dfrac{2\pi \mu_0 (0.02-x)I_1 - 2\pi \mu_0( 0.02+x)I_2 }{ 4\pi^2 (0.02^2-x^2)} =0$$
Thus,
$$ (0.02-x)I_1 - ( 0.02+x)I_2 =0$$
Plug the known;
$$5 (0.02-x) - 3( 0.02+x)= 0.1-5x-0.06-3x=0$$
$$x=\color{red}{\bf 0.5}\;\rm cm$$
$$\color{blue}{\bf [b]}$$
When the two currents are in opposite directions, the magnetic field between the two wires will never be zero at.
$\bullet$ $\bullet$ At $x\gt 0.02$ m, assuming that $I_2$ is into the page while $I_1$ is still out of the page.
$$\sum F_B=0=B_1-B_2$$
$$B_1-B_2=\dfrac{\mu_0I_1}{2\pi d_1}-\dfrac{\mu_0I_2}{2\pi d_2} =0$$
From the last figure below, we can see that at $P_1$, $d_1=x+0.02$, and $d_2=x-0.02$
$$ \dfrac{\mu_0I_1}{2\pi (x+0.02)}-\dfrac{\mu_0I_2}{2\pi (x-0.02)} =0$$
$$ \dfrac{2\pi \mu_0I_1(x-0.02)-2\pi\mu_0I_2 (x+0.02) }{4\pi^2 (x^2-0.02^2)} =0$$
Thus,
$$ I_1 (x-0.02)- I_2 (x+0.02) =0$$
Plug the known;
$$ 5x- 0.1-3x-0.06 =0$$
Thus,
$$x=\color{red}{\bf 8}\;\rm cm$$
$\bullet$ $\bullet$ At $x\lt 0.02$ m, assuming that $I_2$ is into the page while $I_1$ is still out of the page.
$$\sum F_B=0=B_2-B_1$$
$$B_2-B_1=\dfrac{\mu_0I_2}{2\pi d_2}-\dfrac{\mu_0I_1}{2\pi d_1} =0$$
we can see that at $P_1$, $d_2=x'+0.02$, and $d_1=x'-0.02$
$$ \dfrac{\mu_0I_2}{2\pi (x+0.02)}-\dfrac{\mu_0I_1}{2\pi (x-0.02)} =0$$
$$ \dfrac{2\pi \mu_0I_2(x-0.02)-2\pi\mu_0I_1 (x+0.02) }{4\pi^2 (x^2-0.02^2)} =0$$
Thus,
$$ I_2 (x-0.02)- I_1 (x+0.02) =0$$
$$ 3(x-0.02)- 5(x+0.02) =0$$
$$ 3 x-0.06- 5x+-0.1 =0$$
$$x=-8\;\rm cm$$
which is not real since $x$ is measured as distance not position.
Hence, no point at $x\lt -2$ at which the net magnetic field is zero. This is because the field due to the 5.0 A current is always larger than the field due to
the 3.0 A current in this region.
