3. Find all values of the parameter $a$, for which the interval ( $3 a ; 5 a-2)$ contains at least one integer.

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# Solution.

The left end of the interval must be less than the right, so

$3 a < 5a - 2$. Further, if the length of the interval is greater than one, it definitely contains an integer. $5 a - 2 - 3 a > 1 \Rightarrow a > 1.5$. Thus, the interval $(1.5 ; \propto)$ is part of the solution set. Consider the values $a \in (1 ; 1.5]$. For these, the inequalities will hold: 
\[
\left\{\begin{array}{l}
3a - 2 < 3a \\
5a - 2 > 3a
\end{array} \Leftrightarrow \left\{\begin{array}{l}
1 < a \\
a > 1.2
\end{array} \Leftrightarrow a \in (1.2 ; 4 / 3) ;\right.\right.
\]

\[
\left\{\begin{array}{l}
3a - 2 < 5a - 2 \\
5a - 2 > 3a
\end{array} \Leftrightarrow \left\{\begin{array}{l}
1 < a \\
a > 1.4
\end{array} \Leftrightarrow a \in (1.4 ; 1.5]\right.\right.
\]

Combining the found sets, we get: $a \in (1.2 ; 4 / 3) \cup (1.4 ; +\infty)$.

Answer: $(1.2 ; 4 / 3) \cup (1.4 ; +\infty)$.

Remarks on checking. Points 4 and 5 were not considered - 1 point, one of the specified points was considered - 2 points.