This is more of a long comment than an answer.
The "right" notion of an unbounded set being measurable in ZF is less than clear. Suppose $\mathbb{R}$ is a countable union of countable sets. Let $\langle X_n: n<\omega \rangle$ be a partition of $[0, 1]$ into countable sets. Consider $X=\{x+m: n\le m<\omega, x \in X_n\}.$ It's not hard to show that $X$ has infinite outer measure and zero inner measure. Also, responding to a discussion in the comments, $\{\frac{1}{x}: x \in X\}$ is null since it can be written as a union of a countable set and a set of arbitrarily small outer measure. While this set satisfies the definition of measurability Emil provided in the comments, it violates almost everything we expect measurable sets to satisfy.
I think a better definition would be that $X \subset \mathbb{R}$ is measurable if for all $\epsilon>0,$ there is an open cover $C$ of $X$ such that for all bounded intervals $I,$ the restriction of $C$ to $I$ has measure less than $\lambda_*(X \cap I)+\epsilon,$ though I don't know if there's any literature studying this. Anyway, under this definition, your first question has an affirmative answer as demonstrated by the previous example, and the second question seems nontrivial.
Edit: Actually we need both inner and outer uniformity for this to be a finitely additive measure. Say $X$ is measurable if for all $\epsilon>0,$ there are open sets $U_1, U_2$ such that $X \subset U_1,$ $\mathbb{R} \setminus X \subset U_2,$ and $\lambda(U_1 \cap U_2)<\epsilon.$