No. For example, $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ has degree $\phi(n)$, but we usually don't have $\zeta_n^{\phi(n)} \in \mathbb{Q}$. For example, when $n=3$, then $\phi(n)=2$, $\zeta_3 = \frac{-1+\sqrt{-3}}{2}$ and $\zeta_3^2 = \frac{-1-\sqrt{-3}}{2}$.
No. For example, $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ has degree $\phi(n)$, but we usually don't have $\zeta_n^{\phi(n)} \in \mathbb{Q}$. For example, when $n=3$, then $\phi(n)=2$, $\zeta_3 = \frac{-1+\sqrt{-3}}{2}$ and $\zeta_3^2 = \frac{-1-\sqrt{-3}}{2}$.