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The answer is Yes for the two questions. Below the proposed method.

Let $k=\mathbb{F}_2$ be the usual galois field with two elements
integers, we consider

1. the alphabet $X=\{x,y\}$
2. the set of noncommutative series $k<<X>>>=k^{X^*}$ (i.e. all functions $X^*\rightarrow k$ with the convolution product)
3. the augmentation character $k<<X>>>\rightarrow k$ and its kernel $\frak{M}$ (series without constant term)
4. the free group with two generators $\Gamma=\Gamma(x,y)$
5. the group morphism $\mu : \Gamma(x,y)\rightarrow 1+\frak{M}$ $$ \mu(x)=1+x\ ;\ \mu(y)=1+y $$ which is known to be into (Magnus transformation)
6. the quotients $\mathcal{A}_n=k<<X>>>/\frak{M}^n$ (which are finite) and the sujective quotient morphisms $q_n : k<<X>>>\rightarrow \mathcal{A}_n$
7. the groups $\Gamma_n$, images of $q_n\circ \mu$ which are finite

you see that, if your sentence $(\forall x)(\exists y)(w(x,y)=1)$ is true in all finite group, it must be true in all $\Gamma_n$ and then in $\Gamma$, whence in all groups (same for the other sentence).