\( \frac{1}{8}+\frac{3}{8}= \) b) \( \frac{4}{3}-\frac{5}{4}= \) \( \frac{5}{5}+7= \) \( \frac{4}{5}+\frac{7}{5}-\frac{6}{5}= \) \( \frac{7}{4}-\frac{5}{6}= \)

Let's tackle these fractions step by step! First up: \( \frac{1}{8}+\frac{3}{8} = \frac{4}{8} = \frac{1}{2} \). Easy peasy! Now for \( \frac{4}{3}-\frac{5}{4} \). To subtract these, we need a common denominator, which is 12. So \( \frac{4}{3} = \frac{16}{12} \) and \( \frac{5}{4} = \frac{15}{12} \). Thus, \( \frac{16}{12} - \frac{15}{12} = \frac{1}{12} \). Next: \( \frac{5}{5}+7 = 1 + 7 = 8 \). Simple addition! Moving on to \( \frac{4}{5}+\frac{7}{5}-\frac{6}{5} \). Here we can combine like terms easily: \( \frac{4}{5} + \frac{7}{5} - \frac{6}{5} = \frac{4 + 7 - 6}{5} = \frac{5}{5} = 1 \). Finally, for \( \frac{7}{4}-\frac{5}{6} \), use the common denominator of 12 again. So, \( \frac{7}{4} = \frac{21}{12} \) and \( \frac{5}{6} = \frac{10}{12} \). Thus, \( \frac{21}{12} - \frac{10}{12} = \frac{11}{12} \). So, the answers to your calculations are: 1) \( \frac{1}{2} \) 2) \( \frac{1}{12} \) 3) \( 8 \) 4) \( 1 \) 5) \( \frac{11}{12} \)