Let $w$ be a group word with two variables $x$ and $y$.
Is the sentence $(\forall x)(\exists y)w=1$
true in every group if it is true
in every finite group?
The same question about the sentence $(\exists x)(\forall y)w=1$.



On sentences true in all finite groups
The answer is Yes for the second question, about $(\exists x)(\forall y)w=1$. Following Christian Remling's idea:
If a sentence like
$$\exists x(\forall y)(yxy^{-1}x^2y^{-9}\dots=1)$$
holds in all finite groups then it holds in $\mathbb Z/n\mathbb Z$
where it just says (for certain constants $a,b,c,d$)
$$
(\exists x)(\forall y)((a-b)x+(c-d)y=0).
$$
The only way this can be true is if $c=d$.
So *the exponents of $y$ in $w$ add up to 0*.
In that case, the sentence is true in all groups because we can take $x=e$, the group identity (called 1 by the OP).


The answer is also Yes on Question 1. If $\forall x\exists y (w=1) $ holds in 
$\mathbb Z/n\mathbb Z$ then there it says $ ax=by $, i.e., $ b $ divides all $ ax $, so $ b $ divides $ a $. But then in any group given $ x $ we can take $ y=x^{-a/b} $.


On the other hand, [Wikipedia](http://en.m.wikipedia.org/wiki/List_of_first-order_theories#Groups) gives the following $\Pi^0\_2$ sentence where the answer is No:
*given two elements of order 2, either they are conjugate or there is a non-trivial element commuting with both of them*.


