1. Problem: In the figure with circle and tangents, AC is tangent to the circle, BE is parallel to CD, and given angles are (a) $\angle ABE = 42^\circ$, $\angle BDC = 59^\circ$. Find $\angle BED$. (b) $\angle DBE = 62^\circ$, $\angle BCD = 56^\circ$. Find $\angle BED$. 2. Problem: In the figure with circle and tangent PR, find (a) Given $\angle PQT = 66^\circ$, find $\angle QST$. (b) Given $\angle QTS = 38^\circ$ and $\angle QRS = 30^\circ$, find $\angle QST$. **Solution:** 1(a). 1. Since AC is tangent at C, by the tangent-secant theorem, $\angle ABE = \angle BCD$ (alternate segment theorem). 2. Given $\angle ABE = 42^\circ$. 3. Given $\angle BDC = 59^\circ$. 4. Since BE//CD, corresponding angles $\angle BED = \angle BDC = 59^\circ$. Answer: $\boxed{59^\circ}$. 1(b). 1. Given $\angle DBE = 62^\circ$, $\angle BCD = 56^\circ$. 2. By alternate segment theorem, $\angle DBE = \angle DCB$ (tangent AC and chord BC). 3. We want $\angle BED$ which is an exterior angle of triangle BEC. 4. Since BE//CD, $\angle BED = \angle BCD = 56^\circ$. Answer: $\boxed{56^\circ}$. 2(a). 1. PR is tangent to the circle; $\angle PQT = 66^\circ$ is given. 2. By alternate segment theorem, $\angle PQT = \angle QST$. Answer: $\boxed{66^\circ}$. 2(b). 1. Given $\angle QTS = 38^\circ$, $\angle QRS = 30^\circ$. 2. In triangle QRS, sum of angles is $180^\circ$, so $$ \angle QSR = 180^\circ - 38^\circ - 30^\circ = 112^\circ. $$ 3. $\angle QST$ subtends the same chord as $\angle QSR$. 4. By properties of circle and tangent, $\angle QST = 180^\circ - \angle QSR = 68^\circ$. Answer: $\boxed{68^\circ}$.