Problem: Let $A$ and $B$ be two events such that
\[
P(A) = \frac{3}{4} \text{ and } P(B) = \frac{5}{8}.
\]
Show that
\[
P(A \cup B) \geq \frac{3}{4} \text{ and } \frac{3}{8} \leq P(A \cap B) \leq \frac{5}{8}.
\]
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,
\[
\textcolor{blue}{\boxed{
\frac{3}{8} \leq P(A \cap B) \leq \frac{5}{8}
}}
\]