The first half is just a special case of the Szemeredi-Trotter Theorem, which states that for $n$ points and $m$ lines in the plane, the number of incidences is $ \mathcal{O} (n^{\frac{2}{3}}m^{\frac{2}{3}} + m + n) $. Taking $m = n$ and absorbing the linear terms gives the desired conclusion. On the other hand, we can find a configuration with $ \Omega (n^{\frac{4}{3}}) $ incidences. Suppose $n = 8a^3$, and consider choosing all points $(x, y) \in \mathbb{N}^2$ with $x \le a$ and $y \le 8a^2$. Now pass lines of slope $1, 2, \dots, a$ through each point of the form $(1, y)$ where $1 \le y \le a^2$. We can now include $7a^3$ random lines so that the number of points and lines is equal. Notice that each non-random line intersects exactly $a$ points in the grid, so that the total number of incidences is (at least) \[a(a^3) = a^4 = \frac{1}{16}(8a^3)^{\frac{4}{3}} = \frac{1}{16}n^\frac{4}{3} \in \Omega (n^{\frac{4}{3}})\] as desired.