The answer to Q1 is yes, here is a somewhat pedestrian proof.

I'll assume $m,n \geq 2$ are coprime and squarefree, and that $m^{1/a}$ (resp. $n^{1/b}$) denotes the unique real $a$-th root of $m$ (resp. real $b$-th root of $n$).

Recall that $K=\mathbf{Q}(n^{1/b})$ has degree $b$ over $\mathbf{Q}$ (use Eisenstein's criterion). Let us prove directly that $X^a-m$ is irreducible over $K$. Over $\mathbf{C}$ we have

\begin{equation*}
X^a-m = \prod_{k=0}^{a-1} X-\zeta_a^k m^{1/a}
\end{equation*}
If $P$ is a nontrivial factor of $X^a-m$ over $K$, of degree $1 \leq d \leq a-1$, then the constant term of $P$ is of the form $\zeta \cdot m^{d/a}$ for some root of unity $\zeta$. Since $K \subset \mathbf{R}$, we must have $\zeta=\pm 1$, so that $m^{d/a} \in K$. Since $d/a$ is not an integer, we deduce that there must exist a prime $p$ (dividing $a$) such that $m^{1/p} \in K$. In particular $p$ divides $b$.

Now using the Galois correspondence, it is not too hard to show that the unique subfield of degree $p$ of $K$ is $\mathbf{Q}(n^{1/p})$. So we get $\mathbf{Q}(m^{1/p})=\mathbf{Q}(n^{1/p})$. It remains to compare the discriminants of these number fields to get a contradiction. The prime $p$ divides at most one of the numbers $m$ and $n$, say $p$ doesn't divide $m$. Then any prime divisor $q$ of $m$ ramifies in $\mathbf{Q}(m^{1/p})$ but not in $\mathbf{Q}(n^{1/p})$, whence a contradiction.

As noted in my comment, the answer to Q2 is always yes, and one doesn't need the assumption "$m,n$ coprime and squarefree" for that.