1. Let the events be:

   • $P$: person $P$ hits the target, with probability $P(P)=\frac{3}{4}$
   • $Q$: person $Q$ hits the target, with probability $P(Q)=\frac{1}{2}$
   • $R$: person $R$ hits the target, with probability $P(R)=\frac{5}{8}$

   Since they try independently, their miss probabilities are: $P(P')=1-\frac{3}{4}=\frac{1}{4}, \quad P(Q')=1-\frac{1}{2}=\frac{1}{2}, \quad P(R')=1-\frac{5}{8}=\frac{3}{8}$

2. We need the probability that the target is hit by $P$ or $Q$ but not by $R$.

   This means: $(P \cup Q) \cap R'$

3. Because $R$ is independent of $P$ and $Q$, we can write: $P\big((P \cup Q) \cap R'\big)=P(P \cup Q)\cdot P(R')$

4. First compute $P(P \cup Q)$: $P(P \cup Q)=P(P)+P(Q)-P(P\cap Q)$ Since $P$ and $Q$ are independent, $P(P\cap Q)=P(P)P(Q)=\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}$ Therefore, $P(P \cup Q)=\frac{3}{4}+\frac{1}{2}-\frac{3}{8}$ $=\frac{6}{8}+\frac{4}{8}-\frac{3}{8}=\frac{7}{8}$

5. Now multiply by $P(R')=\frac{3}{8}$: $P\big((P \cup Q) \cap R'\big)=\frac{7}{8}\cdot\frac{3}{8}=\frac{21}{64}$

6. Checking options:

   • A: $\frac{21}{64}$ ✅
   • B: $\frac{9}{64}$
   • C: $\frac{15}{64}$
   • D: $\frac{39}{64}$

Therefore, the correct answer is: $\boxed{\frac{21}{64}}$ which corresponds to Option A.