A nice question.
The answer to (1) is No. You have a global field over the constant field $\Bbb F_p$, and I’ll call it $K$ instead of $F$ if you don’t mind. In adjoining a (the) root of the inseparable polynomial $f(Y)=Y^p-a$, you are adjoining the $p$-th root of $a$, and your field is a (pure) inseparable extension of degree $p$, and thus is $K^{1/p}$. In asking whether an element of any completion of $K$ could be a (the) root of $f$, you are asking whether $a$ itself is a $p$-th power in any completion of $K$. In other words, you are asking whether, in a completion $K_{\mathcal P}$, $a$ can possibly be a $p$-th power, in other words an element of $(K_{\mathcal P})^p$.
We may as well agree what a field $K_{\mathcal P}$ looks like: it’ll be Laurent series in a uniformizer $t$, but with coefficients from the
residue field of $\mathcal P$, necessarily a finite extension of $\Bbb F_p$, and so some finite field $\kappa$. Of course $t$ may be taken to be an element of $K$. Now consider the two inseparable extensions $K\supset K^p$ and $K_{\mathcal P}\supset(K_{\mathcal P})^p$, both of degree $p$. For the latter extension, $\{1,t,t^2,\cdots,t^{p-1}\}$ is
certainly a good basis, and indeed it’s a good basis for $K$ over $K^p$ as well, since $t\notin K^p$. Since $a\notin K^p$, when we express it as a linear combination of the basis elements $t^i$ with coefficients in $K^p$, say $$a=\sum_{i=0}^{p-1}\alpha_it^i\,,$$
at least one of the $\alpha_i$ with $1\le i<p$ must be nonzero (otherwise we would have $a\in K^p$).
Now, just as $K_{\mathcal P}=\kappa((t))$, so $(K_{\mathcal P})^p=\kappa((t^p))$, using perfectness of $\kappa$. The $\alpha_i$ from above are also in $(K_{\mathcal P})^p$, so that my displayed equation above can also be read as a statement in the local fields. But notice that $\alpha_it^i$ is a series in which all terms are of degree $\equiv i\pmod p$, because $\alpha_i\in\kappa((t^p))$ and its series has all terms of degree $\equiv0\pmod p$. Conclusion? Since some $\alpha_i$ with $1\le i<p$ is nonzero, the $t$-series for $a$ is not a $p$-th power. That answers your first question negatively.
For your second question, I’m not sure I can give a satisfactory answer. It seems to me that, depending on $a$, you may have elements of at least one completion very close to $a^{1/p}$, simply because you may have elements of $K$ very close to $a^{1/p}$. Indeed, suppose that $K=\Bbb F_p(t)$, and use the $t$-adic valuation. In case $a=1+t^N$ for $N$ a large integer prime to $p$. Then at that particular valuation, $v_t\bigl((1+t^N)-1\bigr)=N$ and $v_t\bigl((1+t^N)^{1/p}-1\bigr)=N/p$, which may be as big as you like.
I hope this was of some help.
