Let $R$ be a local ring with field of fractions $K$, maximal ideal $\mathfrak{m}$ and residue field $\kappa = R/\mathfrak{m}$. Let $R^\mathrm{sh}$ be a strict henselization of $R$, and let $L$ be the field of fractions of $R^\mathrm{sh}$.
(Assuming $R$ is not complete)
For 2, the answer is no, as suggested in a now-deleted comment of LSpice.
For $k$ of characteristic not $2$, with $R$ the localization of $k[t]$ at $0$, the element $\frac{1}{ \sqrt{1+t}+1 }$ is in the strict henselization but not integral. This is the root of the polynomial $$(1-x)^2 - x^2(1+t) = 1 - 2x - t x^2,$$ which reduces mod $t$ to a polynomial with distinct roots and thus has a root in the henselization but which is clearly not divisible by any monic polynomial.
For 1, I think the answer depends on what you mean by "that are unramified at $\mathfrak m$". Recall that an extension of fields may have multiple places lying over $\mathfrak m$, with some ramified and others unramified. You have to choose a lift of $\mathfrak m$ to every extension contained in the separable closure in a consistent way, i.e. choose an extension of the $\mathfrak m$-adic valuation to the separable closure, and then consider the union of the unramified extensions.
The point being that every étale extension of (one-dimensional) local rings comes with a choice of a lift of a valuation (i.e. the one associated to the maximal ideal of the extended ring) and is unramified for this choice, and conversely if you have a field extension with an unramified choice of lift that gives an étale ring extension.
On the other hand, if $R$ is complete (i.e. already non-strictly henselian) then every unramified extension is associated to a separable extension of the residue field and thus has a unique lift of the maximal ideal and is integral, so the answer to both questions is yes.
