I put the following as answer as it does not fit within the limits of a comment, hoping that it could help. I suspect the answer will be YES for the two questions. In fact, the free group is the limit of finite truncations. Below is the proposed method.

Let $k=\mathbb{F}_2$ be the usual galois field with two elements  
integers, we consider 
<ol>
<li> an infinite alphabet $X$ (denumerable is enough)
<li> the set of noncommutative series $\mathcal{A}=k<<X>>>=k^{X^*}$ 
(i.e. all functions $X^*\rightarrow k$ with the convolution product)
<li> the augmentation character $k<<X>>>\rightarrow k$ and 
its kernel $\frak{M}$ (series without constant term) and, for every finite subalphabet $\mathrm{F}\subset X$, the ideal $\frak{M}_\mathrm{F}$ of the series such that every monomial of the support contains at least a letter outside $\mathrm{F}$   
<li> the free group over $X$, $\Gamma=\Gamma(X)$
<li> the group morphism $\mu : \Gamma\rightarrow 1+\frak{M}$ given by
$$
(\forall x\in X)(\mu(x)=1+x)
$$
which is known to be into (Magnus transformation)
<li> the quotients $\mathcal{A}_{n,\mathrm{F}}=k<<X>>>/(\frak{M}^n+ \frak{M}_\mathrm{F})$ (which are finite) 
and the sujective quotient morphisms $q_{n,\mathrm{F}} : k<<X>>>\rightarrow \mathcal{A}_{n,\mathrm{F}}$
<li> the groups $\Gamma_{n,\mathrm{F}}$, images of $q_{n,\mathrm{F}}\circ \mu$ which are finite   
</ol> 
... and if a word $w$ in the free group fails to be $1$ iff it fails to be $1$ in one of the finite groups $\Gamma_{n,\mathrm{F}}$.