Solution: We recall that for given set $A$ and $B$ if $A \subseteq B$, then the corresponding probabilities are $P(A) \leq P(B)$. So, we have \begin{align*} A \subseteq A \cup B & \implies P(A) \leq P(A \cup B) \\ & \implies \frac{3}{4} \leq P(A \cup B). \end{align*} This proves that \[ \textcolor{blue}{\boxed{ P(A \cup B) \geq \frac{3}{4} }} \]

We also recall that for any two events $A$ and $B$, we have \[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \] Using this, we get \[ P(A \cup B) = \frac{3}{4} + \frac{5}{8} - P(A \cap B). \] Since $P(A \cup B) \leq 1$, we have \[ \frac{3}{4} + \frac{5}{8} - P(A \cap B) \leq 1 \implies P(A \cap B) \geq \frac{3}{8}. \] Also, since $A \cap B \subseteq B$, we have $P(A \cap B) \leq P(B)$, we have \[ P(A \cap B) \leq \frac{5}{8}. \] Thus, \[ \textcolor{blue}{\boxed{ \frac{3}{8} \leq P(A \cap B) \leq \frac{5}{8} }} \]