a) If $A$ is a finite Boolean ring, we have for every $x \in A$ $$x+x=(x+x)^2=x^2+x^2+x^2+x^2=x+x+x+x \Longrightarrow x+x= 0.$$Hence $A$ has a vectorspace structure over $\mathbb{F}_2$, and thus of the form $\mathbb{F}_2^k$ for some positive integer $k$. In particular $|A|=2^k$. It is clear that these rings work. b) So we need to show that there exists a countably infinite Boolean ring. An example is the countable direct sum $\bigoplus_{m=1}^\infty \mathbb{F}_2$.

As a remark, in Romania we use the convention in high school mathematics that a ring always has unity. I forgot about this caveat when translating the problem. With this proviso, the direct sum $\bigoplus_{m=1}^\infty \mathbb{F}_2$ does not work, but the direct product $\prod_{m=1}^\infty \mathbb{F}_2$ does. @below: Indeed, nice catch!

Here's a Lagrange's theorem approach. We use the fact proved by @Gryphos that $x + x = 0, \forall x \in A$. Proof. (Only if) Let $|A| = 2^k \cdot m$ for some $k, m \in \mathbb{N}, m > 1$ and $2$ does not divide $m$. Since $2^k > k$, this implies we can find $k$ distinct elements in the ring A, say $x_1, .., x_{k}$. Look at $A' := \{0, x_1, .., x_{k}, x_1 + x_2, .., x_{k-1} + x_{k}, .., x_1 + x_2 + .. + x_{k}\}$ i.e., all $\binom{n}{r}$ "sum" - terms from $r = 1, .., k$. We can easily show that the set $A'$ is an additive group by an inductive approach. This shows that $A'$ is an additive subgroup of $A$. Since $|A'| = 2^k$ and as $m>1$, this implies there is a $x_{k+1}$ distinct from $x_{1}, .., x_{k}$ and the group constructed using $A' \cup \{x_{k+1}\}$ (inductively), will have cardinality $2^{k+1}$ thereby creating a contradiction. $\square$ The second one, it cannot be a direct product as the cardinality is uncountable.