Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.
Now let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.
A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9