The two inequalities intersect on the boundary at $x=2$, $y=\pm \sqrt{10}$, and the ellipse meets the $x$-axis at $x=\pm 2\sqrt{2}$. Enumerating integer points $(x,y)$ satisfying both strict inequalities, the candidates fall on the vertical lines $x=1$ and $x=2$:

$x=1$: $y\in \{-2,-1,0,1,2\}$ (five points).

$x=2$: $y\in \{-3,-2,-1,0,1,2,3\}$ (seven points).

So $|X|=12$ and the sample space size is $\left(\genfrac{}{}{0ex}{}{12}{3}\right)=220$.

Every triangle formed by three of these points has two vertices on a common vertical line $x=1$ or $x=2$ (since any three points include at least two on the same line by pigeonhole). Take the base on that line; the third vertex lies on the other line, so the perpendicular distance (horizontal) is $1$. Hence the area equals $\frac{1}{2}\cdot \text{base}\cdot 1=\frac{\text{base}}{2}$, which is a positive integer iff the base length is an even positive integer.

Count triangles by even base lengths (with base on $x=1$ or $x=2$ and third vertex on the opposite line):

Base $=2$: on $x=1$ there are $3$ such pairs and on $x=2$ there are $5$; combined with $7$ or $5$ choices of the third vertex, gives $3\cdot 7+5\cdot 5=46$.

Base $=4$: $1\cdot 7+3\cdot 5=22$.

Base $=6$: $0\cdot 7+1\cdot 5=5$.

Total favourable triangles $=46+22+5=73$.

Required probability $=\frac{73}{220}$.