Find the sum. \( \begin{array}{ll}\text { 1. } \frac{1}{2}+\frac{2}{3}= & \text { 2. } \frac{8}{12}+\frac{8}{11}= \\ \text { 3. } \frac{2}{7}+\frac{6}{10}= & \text { 4. } \frac{1}{6}+\frac{6}{11}= \\ \frac{5}{9}+\frac{1}{2}= & \text { 6. } \frac{9}{12}+\frac{2}{12}= \\ \frac{2}{7}+\frac{1}{4}= & \text { 8. } \frac{1}{4}+\frac{6}{8}=\end{array} \)

Let's calculate each of these pairs step by step: 1. \( \frac{1}{2} + \frac{2}{3} = \frac{3}{6} + \frac{4}{6} = \frac{7}{6} \) 2. \( \frac{8}{12} + \frac{8}{11} = \frac{2}{3} + \frac{8}{11} = \frac{22}{33} + \frac{24}{33} = \frac{46}{33} \) 3. \( \frac{2}{7} + \frac{6}{10} = \frac{2}{7} + \frac{3}{5} = \frac{10}{35} + \frac{21}{35} = \frac{31}{35} \) 4. \( \frac{1}{6} + \frac{6}{11} = \frac{11}{66} + \frac{36}{66} = \frac{47}{66} \) 5. \( \frac{5}{9} + \frac{1}{2} = \frac{10}{18} + \frac{9}{18} = \frac{19}{18} \) 6. \( \frac{9}{12} + \frac{2}{12} = \frac{11}{12} \) 7. \( \frac{2}{7} + \frac{1}{4} = \frac{8}{28} + \frac{7}{28} = \frac{15}{28} \) 8. \( \frac{1}{4} + \frac{6}{8} = \frac{1}{4} + \frac{3}{4} = 1 \) So the sums are: 1. \( \frac{7}{6} \) 2. \( \frac{46}{33} \) 3. \( \frac{31}{35} \) 4. \( \frac{47}{66} \) 5. \( \frac{19}{18} \) 6. \( \frac{11}{12} \) 7. \( \frac{15}{28} \) 8. \( 1 \)