Given that AB//CD, angle CEF = 5a + 40° and EF B = a + 100°. What is the value of a? State reasons for you answer.

Angles $\angle CEF$ and $\angle EFB$ are consecutive interior angles (also known as co-interior angles) because they are between the parallel lines $AB$ and $CD$ and on the same side of the transversal line $EF$.
When two parallel lines are intersected by a transversal, consecutive interior angles are supplementary, meaning their sum is $180^\circ$.

The value of $a$ is $\boxed{\frac{20}{3}}$.

Angles $\angle PQR$ and $\angle SQR$ are angles on a straight line $PQR$. Therefore, their sum is $180^\circ$.

Since $TS \parallel PR$ and $SQ$ is a transversal, $\angle TSQ$ and $\angle SQR$ are alternate interior angles.

Now, consider the transversal $SR$. Since $TS \parallel PR$, $\angle TSR$ and $\angle SRQ$ are alternate interior angles.

From the diagram, $\angle TSR$ is composed of $\angle TSQ$ and $\angle QSR$.

So, we have the equation:

The sum of angles in a triangle is $180^\circ$.

Substitute the value of $c = 26^\circ$:

Substitute equation (1) into equation (2):

Now substitute the value of $a$ back into equation (1) to find $b$:

(Note: The angle $48^\circ$ is extraneous information not needed to solve for $a, b, c$ based on the standard geometric theorems and parallel line properties.)

The values are:
$a = \boxed{64^\circ}$
$b = \boxed{90^\circ}$
$c = \boxed{26^\circ}$