Using both of these, we conclude that the square elements in $G$ commutes, i.e, $a^2\cdot b^2 = b^2\cdot a^2$.
Now, $(ab)^3 = (ab)\cdot(ab)^2 = (ab)\cdot(b^2\cdot a^2) = (ab)\cdot(a^2\cdot b^2) = a^3\cdot b^3 = a\cdot(a^2 * b^2)\cdot b$, after elimination by right and left cancellation laws, we reckon that the square elements of $G$ commutes with any other elements of $G$, i.e. $a^2\cdot b = b\cdot a^2$.
Since $G$ is 3-abelian, every 3rd power and 2nd power commutes, i.e, $a^3\cdot b^2 = a^2\cdot b^3$. Since $G$ is also 5-abelian, $a^5\cdot b^4 = b^4\cdot a^5$.
It can also be derived that $a^4\cdot b^3 = b^3\cdot a^4$ by the previous property that the square element commutes with any other element of $G$.
so $(ab)^5 = a^5b^5 $ implies $(ba)^4 = a^4b^4$ (1)
and $(ab)^3 = a^3b^3 $ implies $(ba)^2 = a^2b^2$ (2)
so $baba*(ba)^2 = baba*aabb$
using (1) $baba*aabb = a^4b^4$, so
using (2), $aabb*aabb = a^4b^4$or
$bbaa = a^2b^2$
plugging this into (2)
$baba = bbaa$
or ab = ba.