This is what i tried-
$\vert G \vert=2^33^25$. So,
$\vert A \vert=8$ and $\vert B \vert=9$
Now let $a\in A \cap B$ then $ a\in A$ and $a \in B$. Using Lagrange's theorem this implies that $\vert a \vert \mid 8 $ and $\vert a \vert \mid 9 $. This implies that $\vert a \vert=1 $ and since $e$ is the only element whose order is 1, we have $A\cap B=\{e\}$. So $\vert A\cap B \vert=1$.
Please let me know if this is a correct conclusion.
Your proof is correct but because you are using Lagrange's Theorem which is result about cardinality of subgroup, it will be better if you consider $|A\cap B|$ rather than $|a|$. Of course your proof is not wrong because $|a|=|\langle a\rangle|$, where $\langle a\rangle$ is a subgroup.
Since $A\cap B$ is a subgroup of $A$ and $B$, by Lagrange's Theorem, $|A\cap B|$ divides $|A|$ and $|B|$ and hence divides $\gcd(|A|,|B|)$, which is 1 since $|A|$ and $|B|$ are relatively prime.
Hence $|A\cap B|=1$.