Answer: 244

Small Hint:

Each choice of two $A$’s and two $B$’s produces exactly one interior crossing, so there are $\binom{7}{2}\binom{5}{2}$ crossing points

Big Hint:

Use Euler’s formula $V - E + F = 2:$ each crossing splits two segments, so the segments contribute $mn + 2\binom{m}{2}\binom{n}{2}$ edges

Solution:

Two segments $\overline{A_iB_j}$ and $\overline{A_kB_l}$ cross strictly between the lines exactly when one of the $A$’s comes first and the other’s $B$ comes first, which happens for exactly one pairing of any two $A$’s with any two $B$’s. By the general-position hypothesis these crossings are distinct, so there are $X = \binom{m}{2}\binom{n}{2}$ of them.

Clip the two lines to long segments and apply Euler’s formula. The vertices are the $m + n$ marked points, the $X$ crossings, and the $4$ clipped line ends, so $V = m + n + X + 4.$ Line $\ell_A$ is divided into $m + 1$ edges and $\ell_B$ into $n + 1;$ each crossing splits two segments, so the drawn segments contribute $mn + 2X$ edges, giving $E = mn + m + n + 2X + 2.$ Then $$F = E - V + 2 = mn + X,$$ of which one face is unbounded, so there are $mn + X - 1$ bounded regions. For $m = 3,$ $n = 2$ this gives $6 + 3 - 1 = 8,$ matching the figure.

For $m = 7$ and $n = 5:$ $35 + \binom{7}{2}\binom{5}{2} - 1$ $= 35 + 21 \cdot 10 - 1$ $= 244.$