Given ABF = 50^.

The angle between a tangent ($\overline{BE}$) and a chord ($\overline{AB}$) through the point of contact ($B$) is equal to the angle in the alternate segment ($\angle ACB$). This is the Alternate Segment Theorem.
Given $\angle ABF = 50^\circ$.
Therefore,

$$\angle ACB = 50^\circ$$

We are given that $\overline{AC} \parallel \overline{BE}$. Since $\overline{BC}$ is a transversal intersecting these parallel lines, the alternate interior angles are equal.
So, $\angle CBF = \angle ACB$.
We are given $\angle CBF = y^\circ$.
Therefore,

$$y = \angle ACB = 50$$ $$\boxed{y = 50}$$

Since $\overline{AC} \parallel \overline{BE}$ and $\overline{AB}$ is a transversal, the alternate interior angles are equal.
So, $\angle CAB = \angle ABE$.
The angle $\angle ABE$ is the angle between the tangent $\overline{BE}$ and the chord $\overline{AB}$, which is $\angle ABF = 50^\circ$.
Therefore,

$$\angle CAB = 50^\circ$$

Angles subtended by the same arc ($\overline{BC}$) at the circumference are equal. Both $\angle BFC$ and $\angle BAC$ are subtended by arc $\overline{BC}$.
Therefore,

$$\angle BFC = \angle BAC = 50^\circ$$

From the diagram, $D$ is a point on the line $BC$ extended, meaning $B$, $C$, and $D$ are collinear. Thus, $\angle DFC$ is the same angle as $\angle BFC$.
We are given $\angle DFC = x^\circ$.
Therefore,

$$x = \angle BFC = 50$$ $$\boxed{x = 50}$$