In general we will have to integrate. But here the geometry is enough. Draw a picture. (A pcture can also be handy if we need to integrate.)
We have $Y\le 1/2$ if we are in the bottom half of the left-hand square (probability $(1/4)(1/2)$), or if we are in the right-hand square (probability $3/4$), for a total of $7/8$.
Now we can find the cumulative distribution function of $X$, given that $Y\lt 1/2$. Of course this is $0$ if $x\lt -1$, and $1$ if $x \gt 1$. It remains to take care of things when $-1\le x\le 1$.
For $-1\le x\le 0$, we want to calculate $\frac{\Pr((X \le x)\cap (Y\lt 1/2))}{\Pr(Y\lt 1/2)}.$ The numerator is $(1/4)(1/2)(x-(-1))$. Divide by $7/8$ and simplify. We get $\dfrac{x+1}{7}$.
For $0 \lt x \lt 1$, the calculation is similar. We have $\Pr((X \le x)\cap (Y\lt 1/2))=\frac{1}{8}+\Pr(0\le X \le x)=\frac{1}{8}+\frac{3}{4}x.$ Divide by $7/8$. We get $\dfrac{6x+1}{7}$.