By definition, two angles form a linear pair if and only if 

(1) they have the same vertex;

(2) they share a side; and,

(3) their interiors have no points in common.

$\displaystyle \angle CBE$ is the angle with vertex $\displaystyle B$; its two sides are the rays $\displaystyle \overrightarrow{BC}$ and $\displaystyle \overrightarrow{BE}$, which have endpoint $\displaystyle B$ and pass through $\displaystyle C$ and $\displaystyle E$, respectively. $\displaystyle \angle CBD$ has the same vertex; its two sides are the rays $\displaystyle \overrightarrow{BC}$ and $\displaystyle \overrightarrow{BD}$, which have endpoint $\displaystyle B$ and pass through $\displaystyle C$ and $\displaystyle E$, respectively. $\displaystyle \angle CBE$ and $\displaystyle \angle CBD$ are indicated below in red and green, respectively:

The angles have the same vertex and they share a side. However, the interior of $\displaystyle \angle CBE$ is entirely contained in the interior of $\displaystyle \angle CBD$. The angles do not comprise a linear pair.

A quadrilateral is named after its four vertices in consecutive order, going clockwise or counterclockwise. Quadrilateral $\displaystyle BEFD$ is the figure in red, below:

$\displaystyle E$, $\displaystyle F$, $\displaystyle D$, and $\displaystyle B$ also name the four vertices in clockwise order. It follows that Quadrilateral $\displaystyle EFDB$ is another valid name for the figure.

A quadrilateral is named after its four vertices in consecutive order, going clockwise or counterclockwise. Quadrilateral $\displaystyle BEFD$ is the figure in red, below:

$\displaystyle D$, $\displaystyle F$, $\displaystyle E$, and $\displaystyle B$ name the four vertices in counterclockwise order. It follows that Quadrilateral $\displaystyle DFEB$ is another valid name for the figure.

$\displaystyle \angle 1$ and $\displaystyle \angle 2$ form a pair of vertical angles, and are consequently congruent whether or not it holds that $\displaystyle m || n$. Therefore, whether the lines are parallel cannot be determined.

$\displaystyle \angle 2$ and $\displaystyle \angle 3$ are both inside the two lines $\displaystyle m$ and $\displaystyle n$, and they appear on opposite sides of transversal $\displaystyle t$. They are thus alternate interior angles, by definition, and since they are congruent, then by the Converse of the Alternate Interior Angles Theorem, it follows that $\displaystyle m || n$,

 The letters in the name of a triangle name its vertices, so $\displaystyle \bigtriangleup ACF$ refers to the triangle with vertices $\displaystyle A$, $\displaystyle C$, and $\displaystyle F$. A triangle is named after its vertices in any order, so this triangle can also be called $\displaystyle \bigtriangleup A FC$.

$\displaystyle \angle 1$ and $\displaystyle \angle 2$ form a linear pair of angles and are supplementary regardless of whether or not $\displaystyle m || n$. 

$\displaystyle \angle 1$ and $\displaystyle \angle 2$ form a linear pair of angles, and are therefore supplementary; the same holds for $\displaystyle \angle 3$ and $\displaystyle \angle 4$. Angles that are supplementary to congruent angles are themselves congruent, so, since $\displaystyle \angle 1 \cong \angle 4$, it follows that $\displaystyle \angle 2 \cong \angle 3$.

Notice that $\displaystyle \angle4$ and $\displaystyle \angle6$ are corresponding angles. This means that they possess the same angular measurements; thus, we can write the following:

$\displaystyle \angle4=\angle6$

Now, notice that $\displaystyle \angle4$ and the provided angle of $\displaystyle 78^\circ$ angle are vertical angles. Vertical angles share the same angle measurements; therefore, we may write the following:

If $\displaystyle \angle4=\angle6$ and $\displaystyle \angle4=78^\circ$, then $\displaystyle \angle6=78^\circ$

Let's begin by observing the larger angle. $\displaystyle \angle ABC$ is cut into two 10-degree angles by $\displaystyle \overrightarrow{BD}$. This means that angles $\displaystyle \angle ABD$ and $\displaystyle \angle CBD$ equal 10 degrees. Next, we are told that $\displaystyle \overrightarrow{BE}$ bisects $\displaystyle \angle CBD$, which creates two 5-degree angles.  $\displaystyle \angle ABE$ consists of $\displaystyle \angle ABD$, which is 10 degrees, and $\displaystyle \angle DBE$, which is 5 degrees. We need to add the two angles together to solve the problem.

$\displaystyle \angle ABE=\angle ABD+\angle DBE$

$\displaystyle \angle ABE=10^{\circ}+5^{\circ}$

$\displaystyle \angle ABE=15^{\circ}$