for the first part, you have x-w to 1/2. So, you have y=x-w to y=1/2. Where are we getting y=1/2 from? I'm assuming it's from the uniform distribution, but not sure how to calculate it to get it to 1/2.
Ok first off make sure you understand that $W=X-Y = (\theta + U[-1/2,1/2]) - (\theta + U[-1/2,1/2]) = U[-1/2,1/2]-U[-1/2,1/2]$
Further these are independent so the joint distribution of $X,Y$ is a 2D Uniform variate on $[-1/2,1/2]\times [-1/2,1/2]$
We want $P[W < w] = P[X-Y < w]$.
There are two cases, $w< 0,~0\leq w$
I'll show 3 pictures. The first picture shows when $X-Y < w$ and $w<0$
The beige area is the region that satisfies $P[W < -0.75]$
It should be clear that $x$ ranges from $-1/2$ to some value and that $y$ ranges from some value to $1/2$ The line of the boundary is $Y=X - w$
The $x$ limit occurs when $X-w$ equals the max of $Y$ which is $1/2$, so $X-w = 1/2,~X = 1/2+w$
So the integration limits on $x$ are $-1/2 \to w + 1/2$
Similarly we find the lower limit for $Y$ by noting $X - Y < w \Rightarrow Y > X - w$ and thus for a given $x$, $y > x-w$
Thus the limits on $y$ are $x-w \to 1/2$
The second picture is where $X=Y$
It should be pretty clear $P[W \leq 0] = 1/2$
Now look at $w>0$
Here it's easier to look at $1-P[W > w]$
This corresponds to the purple area.
It should be clear that $x$ goes from some value based on $w$ to $1/2$
And likewise $y$ goes from $-1/2$ to some value based on $w$
$X-Y > w \Rightarrow X > w+Y \geq w -1/2$
and thus the limits on $x$ are $w - 1/2 \to 1/2$
$X-Y > w \Rightarrow Y < X-w \leq 1/2 - w$
and thus the limits on $y$ are $-1/2 \to 1/2 - w$
The coefficient of $1/2$ in front of the last integral is wrong. It should just be 1.
Hopefully you can understand all this. This is a pretty standard problem.