On sentences true in all finite groups (revisited)

Let $w$ be a group word with variables $\bar x, \bar y$, where $\bar x=(x_1,\dots ,x_m)$ and $\bar y=(y_1,\dots ,y_n).$ I am interested in the following questions.

(1) Is the sentence $(\forall\bar x)(\exists\bar y)w=1$ true in every group if it is true in every finite group?

(2) Is the sentence $(\exists\bar x)(\forall\bar y)w=1$ true in every group if it is true in every finite group?

For $m=2$ and $n=4$ the answer to (1) is negative, in general, as shown in T. Coulbois, A. Khelif, Proc. AMS 127 (1999), No. 4, 963--965.

In On sentences true in all finite groups I asked the questions (1) and (2) for $m=n=1$.

Bjørn Kjos-Hanssen gave positive answers to these questions. In fact, his answers can be generalized: an answer is positive for (1) if $m=1$, and for (2) if $n=1$. The open questions I want to ask:

(a) Is $(\forall xy)(\exists z)w(x,y,z)=1$ true in every group if it is true in every finite group?

(b) Is $(\exists x)(\forall yz)w(x,y,z)=1$ true in every group if it is true in every finite group?

• This time it will have to involve non-abelian groups it seems. – Bjørn Kjos-Hanssen May 29 '15 at 2:45
• Yes, you are right. Also, I guess that, probably, answers to these questions are negative. – owb May 29 '15 at 3:19
• About (2): The sentence $\exists x\forall y_1,\dots,y_k:w(x,y_1,\dots,y_k)=1$ holds in every group iff it holds in a free group of countable rank, iff $w(1,y_1,\dots,y_k)=1$ as abstract word. I'm not sure if the same conclusion if the above sentence is only supposed to hold in every finite group. – YCor May 29 '15 at 12:57