Choose $n=a^3b^2$ and the rest is easy.

n=a^5 * b^9 also works

Definitely true.We proceed as follows Choose $n=b^{6k+2} \cdot a^{6k+3} \cdot r^2$ Then it follows obviously #Remark In fact we can find $n$ such that $an$ is a perfect square, $a_2n$ such that it is a perfect cube and $a_tn$ is a perfect. $t$ power.

You can also use Chinese Reminder Theorem, let $a=p_1^{x_1}p_2^{x_2}\ldots p_n^{x_n}$ and $b=p_1^{y_1}p_2^{y_2}\ldots p_n^{y_n}$, where $x_i,y_i\geq 0$. Say $n=p_1^{z_1}p_2^{z_2}\ldots p_n^{z_n}$, where $x_i,y_i\geq 0$, therefore we must have by the given conditions that $$x_i+z_i\equiv 0\pmod{2}$$ and $$y_i+z_i\equiv 0\pmod{3}.$$ By Chinese Reminder theorem, there exist unique $z_i\pmod{6}$ for $x_i,y_i$ for all $i$. We are done.