ag.algebraic geometry - Hodge numbers of compactifications

Let $E$ and $F$ be two elliptic curves and let the involution $sigma$ act on
$Etimes F$ by $sigma(e,f)=(-e,f+alpha)$, $alpha$ is an element of order two
of $F$. Finally let $overline{X}=(Etimes F)/sigma$ (this is a so called hyperelliptic
surface). We have an inclusion $F':=0times F/langlealpharanglesubseteq S$ 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 all wrong it took care of $H^3(X)$ but not (the more interesting) $H^1(X)$. At the moment I am less sure than I was that the answer to 1) is no.

As for 2) you can just look at $mathbb A^3subseteqmathbb P^3$ 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).