Please don't judge by pointing to my other questions here, but all I got is that the lower angle is 45 deg. Interior angles sum to 180(n-2), I think, sooo...
$$360 -45 = 180(n-2)-45 = 2x+y$$ where I guess the answer is whatever $y$ is and x is either of the two remaining angles.
If so, how do I proceed?
If not, where I did err, and how might I instead approach this?
The angle where the dart is pointing (you call it the lower angle) has measure $\frac{1}{8} \cdot 360^{\circ} = 45^{\circ}$. This is because, if the darts are arranged in a circular array to make up the star shaped polygon (slice the star into eight darts by the eight long lines of symmetry), the lower angles of all the darts add up to $360^{\circ}$, and there are $8$ such angles.
Moreover, because the star shaped polygon is made of overlapping squares, each outward corner is a right angle. The angles of the back of the dart are formed via bisection of such outward corners. Therefore the two angles of the back of the dart (call them the upper angles for the sake of your picture) each have measure$\frac{1}{2} \cdot 90 = 45^{\circ}$.
Now, the four interior angles of the dart sum to $360^{\circ}$. We've found three interior angles, each of $45^{\circ}$. To follows that the remaining (obtuse) angle of the dart has measure $360^{\circ} - 3 \cdot 45^{\circ} = 225^{\circ}$. This is clearly the largest interior angle of the dart.