1. **Problem Q2:** Given that lines AB and CD are parallel, find the missing angle $y$. 2. **Formula and rules:** When two parallel lines are cut by a transversal, alternate interior angles are equal, and the sum of angles on a straight line is $180^\circ$. 3. **Step 1:** Identify the angles on line AB: $65^\circ$ and $58^\circ$. 4. **Step 2:** Calculate the angle adjacent to $y$ on line CD using the straight line rule: $$65^\circ + 58^\circ + y = 180^\circ$$ 5. **Step 3:** Simplify the sum of known angles: $$65 + 58 = 123$$ 6. **Step 4:** Substitute and solve for $y$: $$123 + y = 180$$ $$y = 180 - 123$$ $$y = 57^\circ$$ 7. **Reason:** The angles on a straight line sum to $180^\circ$. --- 8. **Problem Q3:** Given lines ABC and DEFG are parallel, $BE = EF$, angle at $B$ is $38^\circ$, find $x$ and $y$. 9. **Formula and rules:** In an isosceles triangle (since $BE = EF$), the base angles are equal. Also, the sum of angles in a triangle is $180^\circ$. 10. **Step 1:** Triangle $BEF$ is isosceles with $BE = EF$, so angles at $E$ and $F$ are equal: $$x = y$$ 11. **Step 2:** The sum of angles in triangle $BEF$ is: $$38^\circ + x + y = 180^\circ$$ 12. **Step 3:** Substitute $x = y$: $$38 + x + x = 180$$ $$38 + 2x = 180$$ 13. **Step 4:** Solve for $x$: $$2x = 180 - 38$$ $$2x = 142$$ $$x = 71^\circ$$ 14. **Step 5:** Since $x = y$, then: $$y = 71^\circ$$ 15. **Reason:** Base angles of an isosceles triangle are equal, and the sum of angles in a triangle is $180^\circ$. **Final answers:** - Q2: $y = 57^\circ$ - Q3: $x = 71^\circ$, $y = 71^\circ$