In a circle with center O, points P, Q, and T are on the circumference. If OP = OQ and angle OPQ = 24 degrees, find angles x (angle POQ) and y (angle PTQ).

Since $O$ is the center of the circle, $OP$ and $OQ$ are both radii. Therefore, $\triangle OPQ$ is an isosceles triangle with $OP = OQ$.

In an isosceles triangle, the angles opposite the equal sides are equal. So, $\angle OQP = \angle OPQ$.
Given $\angle OPQ = 24^\circ$, thus $\angle OQP = 24^\circ$.

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

Angle $x$ is $\boxed{132^\circ}$.

Angle $x$ ($\angle POQ$) is the angle subtended by arc $PQ$ at the center. Angle $y$ ($\angle PTQ$) is the angle subtended by the same arc $PQ$ at the circumference.
The angle at the center is twice the angle at the circumference.

Angle $y$ is $\boxed{66^\circ}$.

$SR$ is a tangent to the circle at $Q$. $\angle SQR$ is the angle between the tangent $SR$ and the chord $PQ$.
According to the Alternate Segment Theorem, this angle is equal to the angle in the alternate segment, which is $\angle PTQ$ (angle $y$).

Angle $m$ is $\boxed{66^\circ}$.