75. Find all fractions with single-digit denominators, each of which is greater than $\frac{7}{9}$ but less than $\frac{8}{9}$.
75. We will replace the given fractions with equal ones by multiplying the numerator and denominator by $2,3, \ldots, 8$. As a result, we get:
1) $\frac{14}{18}$ and $\frac{16}{18}$; between these fractions, there is one fraction $\frac{15}{18}$, which, after simplification, gives $\frac{5}{6}$.
2) $\frac{21}{27}$ and $\frac{24}{27}$. We have $\frac{22}{27}$ and $\frac{23}{27}$. This does not satisfy the condition.
3) $\frac{28}{36}$ and $\frac{32}{36}$. We have $\frac{29}{36} ; \frac{30}{36} ; \frac{31}{36}$. Only one fraction $\frac{30}{36}=\frac{5}{6}$ satisfies the condition, but this one already exists.
4) $\frac{35}{45}$ and $\frac{40}{45}$. We have $\frac{36}{45} ; \frac{37}{45} ; \frac{38}{45} ; \frac{39}{45}$. Only one fraction $\frac{36}{45}=\frac{4}{5}$ satisfies the condition.
5) $\frac{42}{54}$ and $\frac{48}{54}$. Out of five fractions, only one $\frac{45}{54}=\frac{5}{6}$ satisfies the condition, but this one already exists.
6) $\frac{49}{63}$ and $\frac{56}{63}$. Out of six fractions, only one satisfies the condition $\frac{54}{63}=\frac{6}{7}$.
7) $\frac{56}{72}$ and $\frac{64}{72}$. Out of seven fractions, two satisfy the condition: $\frac{60}{72}=\frac{5}{6}$ and $\frac{63}{72}=\frac{7}{8}$, but only one is new. Therefore, the required fractions are $\frac{5}{6} ; \frac{4}{5} ; \frac{6}{7}$ and $\frac{7}{8}$.