Let $E$ and $F$ be two elliptic curves and let the involution $\sigma$ act on
$E\times F$ by $\sigma(e,f)=(-e,f+\alpha)$, $\alpha$Let $E$ and $F$ be two elliptic curves and let the involution $\sigma$ act on
$E\times F$ by $\sigma(e,f)=(-e,f+\alpha)$, $\alpha$ is an element of order two
of $F$. Finally let $\overline{X}=(E\times F)/\sigma$ (this is a so called hyperelliptic
surface). We have an inclusion $F':=0\times F/\langle\alpha\rangle\subseteq S$ and
put $X:=\overline{X}\setminus F'$. Then $X$ is Hodge-Tate but all other good
compactifications of $X$ are obtained by blowing ups and downs of $\overline{X}$
which means that you can never get rid of $F'$ (alternatively any good
compactification $X'$ has $H^1(X')=H^1(X)$ and you need something non-Hodge-Tate
at the boundary to kill that off).
Addendum: This example is an elementall wrong it took care of order two of $F$. Finally let $\overline{X}=(E\times F)/\sigma$$H^3(X)$ but not (this is a so called hyperelliptic surfacethe more interesting). We have an inclusion $F':=0\times F/\langle\alpha\rangle\subseteq S$ and put $X:=\overline{X}\setminus F'$$H^1(X)$. Then $X$ is Hodge-Tate but all other good compactifications of $X$ are obtained by blowing ups and downs of $\overline{X}$ which meansAt the moment I am less sure than I was that you can never get rid of $F'$ (alternatively any good compactification $X'$ has $H^1(X')=H^1(X)$ and you need something non-Hodge-Tate at the boundaryanswer to kill that off1) is no.
As for 2) you can just look at $\mathbb A^3\subseteq\mathbb P^3$ which is a good Hodge-Tate compactification with $\mathbb P^2$ as divisor at infinity and then blow up something non-Hodge-Tate in $\mathbb P^2$. This gives a good compactification with two components one of which (the exceptional divisor for the blowing up) is non-Hodge-Tate (as is the intersection of these two divisors).