Answer
See the detailed answer below.
Work Step by Step
At point 1, the directions of the magnetic fields exerted by the two wires are out of the page.
So the net magnetic eclectic field at point 1 is given by
$$\sum B_1=B_1+B_2$$
and since both wires have the same current $5\;\rm A$, and at the same distance from point 1,
$$\sum B_1=2B_1=\dfrac{2\mu_0I}{2\pi d_1}$$
From the geometry of the figure below, we can see that $d_1=0.04\sin75^\circ$
Plug the known;
$$\sum B_1=\dfrac{2(4\pi \times 10^{-7})(5)}{2\pi(0.04\sin75^\circ) }$$
$$\sum B_1=\color{red}{\bf 5.2\times 10^{-5}}\;\rm T\tag{Out of the page}$$
At point 2, the magnetic fields exerted by the two wires are opposing each other.
So the net magnetic eclectic field at point 1 is given by
$$\sum B_2=B_1-B_2$$
and since both wires have the same current $5\;\rm A$, and at the same distance from point 2,
$$\sum B_2=\color{red}{\bf 0}\;\rm T$$
