Solution:

Both right-hand sides take values in $[0, 1],$ so every intersection lies in the unit square, and there we may write both curves using $\varphi(u) = 4\left||u - \tfrac{1}{2}| - \tfrac{1}{4}\right|:$ the first is $y = \varphi(|\sin 2\pi x|)$ and the second is $x = \varphi(|\cos 3\pi y|).$ As $u$ increases from $0$ to $1,$ $\varphi(u)$ runs linearly $1 \to 0 \to 1 \to 0 \to 1$ with corners at $u = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}.$ For $x \in [0, 1],$ $|\sin 2\pi x|$ sweeps $[0, 1]$ monotonically $4$ times, so the first graph consists of $4 \cdot 4 = 16$ monotone arcs, each climbing or descending through the full range $0 \le y \le 1$ within a narrow vertical strip. Likewise $|\cos 3\pi y|$ sweeps $[0, 1]$ monotonically $6$ times for $y \in [0, 1],$ so the second graph consists of $24$ monotone arcs, each crossing the full range $0 \le x \le 1$ within a narrow horizontal strip.

Take one arc of each graph, living in the vertical strip $[a, b]$ and the horizontal strip $[c, d].$ Inside the rectangle $[a, b] \times [c, d],$ the first arc joins the bottom edge to the top edge and the second joins the left edge to the right edge, and each is monotone, so the two arcs cross exactly once. This yields $16 \cdot 24 = 384$ intersection points.

One further point hides at the corner $(1, 1),$ which lies on both graphs: $\varphi(|\sin 2\pi|) = \varphi(0) = 1$ and $\varphi(|\cos 3\pi|) = \varphi(1) = 1.$ Near it the first graph is $y \approx 1 - 8\pi(1 - x)$ while the second satisfies $x \approx 1 - 18\pi^2(1 - y)^2,$ so the two final arcs meet at their shared endpoint $(1, 1)$ in addition to the transversal crossing already counted. The total is $384 + 1 = 385.$