Let $K$ be a complete discrete valuation field of characteristic $(0, p)$ with perfect residue field $k$. Fix $\overline{K}$ an algebraic closure of $K$ and with absolute galois group $G_K$. Denote $\mathcal{O}_{\overline{K}}$ the ring of interger of ${\overline{K}}$. Fix $\pi$ a uniformizer of $K$.
Define $\mathcal{R}:=\varprojlim_{x\mapsto x^p}\mathcal{O}_{\overline{K}}/p\mathcal{O}_{\overline{K}}$, with the transition morphism Frobenius map.
Fix $\tilde{\pi}:=(\pi_n)_{n\geq 0}\in \mathcal{R}$ with $\pi_0=\pi$ and $\varepsilon:=(\zeta_{p^i})_{i\geq 0}\in \mathcal{R}$ with $\{\zeta_{p^i}\} $ a fixed system of primitive $p^i$-th root of unity. Denote $K_{\pi}=\cup_{n}K(\pi_n)$ with galois group $Gal(\overline{K}/K_{\pi})=:G_{\pi}$ and $K_{\zeta}:=\cup_{n}K(\zeta_{p^n})$ with Galois group $G_{\zeta}$, while $L:=K_{\pi}K_{\zeta}$ with galois group $G_L$.
Now we have $k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!):= \cup_n k(\!(\tilde{\pi})\!)((\varepsilon-1)^{1/p^n}) \subset Frac\mathcal{R}$ and we denote its separable closure $k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!)^{sep}$. Notice that $G_K$ acts over $Frac\mathcal{R}$.
Notice that $(k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!)^{sep})^{G_L}\supset k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!)$ since $G_L$ fix $\tilde{\pi}$ and $\varepsilon$.
The questions are:
Is the galois group $Gal(k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!)^{sep}/k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!))$ a subgroup of $G_{\pi}$?
Is the inclusion $(k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!)^{sep})^{G_L}\supset k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!)$ an equality? If not, which are those missed elements?
2.1 Are roots of the polynomial $X^p+\tilde{\pi}X+\tilde{\pi}$ contained in $(k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!)^{sep})^{G_L}$?
(Remark: (1) $\tilde{\pi}^{1/p}$ should not be in $k(\!(\tilde{\pi}, (\varepsilon-1)^{1/p^{\infty}})\!)^{sep}$ as $X^p-\tilde{\pi}$ is not a separable polynomial in characteristic $p$ case.
(2) Denote $\lambda$ one root of $X^p+\tilde{\pi}X+\tilde{\pi}$. Then $\lambda^p=\tilde{\pi}(\lambda+1)$. Notice $u:=1+\lambda$ is a unit ($\lambda$ must have positive naluation) and we have $v_p(\lambda)=1/p$ and $\lambda^p=\tilde{\pi} u$. What can we say about $\lambda$?)