Hexagon $ABCDEF$ is inscribed in a circle.  If $\measuredangle ACE = 35^{\circ}$ and $\measuredangle CEA = 55^{\circ}$, then compute the sum of the degree measures of $\angle ABC$ and $\angle EFA$.

[i]Proposed by Isabella Grabski[/i]
1. Given that hexagon $ABCDEF$ is inscribed in a circle, we know that $\angle ACE = 35^\circ$ and $\angle CEA = 55^\circ$.
2. Since $\angle ACE$ and $\angle CEA$ are inscribed angles, they intercept arcs $\overarc{AE}$ and $\overarc{AC}$ respectively.
3. The measure of an inscribed angle is half the measure of the intercepted arc. Therefore:
   \[
   \angle ACE = \frac{1}{2} \overarc{AE} \implies 35^\circ = \frac{1}{2} \overarc{AE} \implies \overarc{AE} = 70^\circ
   \]
   \[
   \angle CEA = \frac{1}{2} \overarc{AC} \implies 55^\circ = \frac{1}{2} \overarc{AC} \implies \overarc{AC} = 110^\circ
   \]
4. Since the total measure of the circle is $360^\circ$, the remaining arcs $\overarc{CA}$ and $\overarc{EA}$ can be calculated as:
   \[
   \overarc{CA} = 360^\circ - \overarc{AC} = 360^\circ - 110^\circ = 250^\circ
   \]
   \[
   \overarc{EA} = 360^\circ - \overarc{AE} = 360^\circ - 70^\circ = 290^\circ
   \]
5. The angles $\angle ABC$ and $\angle EFA$ are inscribed angles that intercept the longer arcs $\overarc{CA}$ and $\overarc{EA}$ respectively. Therefore:
   \[
   \angle ABC = \frac{1}{2} \overarc{CA} = \frac{1}{2} \times 250^\circ = 125^\circ
   \]
   \[
   \angle EFA = \frac{1}{2} \overarc{EA} = \frac{1}{2} \times 290^\circ = 145^\circ
   \]
6. Adding these two angles together, we get:
   \[
   \angle ABC + \angle EFA = 125^\circ + 145^\circ = 270^\circ
   \]

The final answer is $\boxed{270^\circ}$