1. **Problem statement:** Given a circle with points N, Q, O on the circumference and points P, R outside the circle, we know that $m\angle Q\hat{N}O = 238^\circ$. We need to find $m\angle P\hat{Q}O$ and $m\angle P\hat{Q}R$. 2. **Understanding the angles:** The angle $m\angle Q\hat{N}O = 238^\circ$ is an exterior angle at point N formed by points Q and O on the circle. Since the full circle is $360^\circ$, the interior angle $m\angle QNO$ on the circle is $360^\circ - 238^\circ = 122^\circ$. 3. **Using the circle theorem:** The angle subtended by chord $NO$ at the circumference is half the measure of the central angle subtending the same chord. Since $m\angle QNO = 122^\circ$, the angle subtended by chord $NO$ at point Q on the circumference is half of $122^\circ$, so $$m\angle PQO = \frac{122^\circ}{2} = 61^\circ.$$ 4. **Finding $m\angle PQR$:** Points P, Q, R are collinear with Q outside the circle. The angle $m\angle PQR$ is formed by the intersection of lines through points P, Q, R and N, Q, O. Since $m\angle PQO = 61^\circ$ and $m\angle Q\hat{N}O = 238^\circ$, the angle $m\angle PQR$ is the supplementary angle to $m\angle PQO$ at point Q, so $$m\angle PQR = 180^\circ - 61^\circ = 119^\circ.$$ **Final answers:** $$m\angle PQO = 61^\circ$$ $$m\angle PQR = 119^\circ$$