(a) Yes. Here is an example:
$$A = \{3k \mid k \in \mathbb{Z}\}, \quad B = \{3k + 1 \mid k \in \mathbb{Z}\}, \quad C = \{3k + 2 \mid k \in \mathbb{Z}\}.$$

(b) No. Suppose $\mathbb{Q}$ can be partitioned into three nonempty subsets $A, B, C$ such that $A+B, B+C, C+A$ are pairwise disjoint. Note that for all $a \in A, b \in B, c \in C$ we have
$$a + b - c \in C, \quad b + c - a \in A, \quad c + a - b \in B. \qquad (1)$$
Indeed $a + b - c \notin A$ since $(A+B) \cap (A+C) = \emptyset$, and likewise $a + b - c \notin B$, hence $a + b - c \in C$. That is, $A+B \subset C+C$. Similarly, we have $B+C \subset A+A$, $C+A \subset B+B$.
The reverse inclusion also holds. Let $a, a' \in A, b \in B, c \in C$. By (1) we get $a' + c - b \in B$, and since $a \in A, c \in C$, applying (1) again gives
$$a + a' - b = a + (a' + c - b) - c \in C.$$

$A + B = C + C$, $B + C = A + A$, $C + A = B + B$.
Furthermore, without loss of generality assume $0 \in A$. Then $B = \{0\} + B \subset A + B$ and $C = \{0\} + C \subset A + C$. Since $B + C$ is disjoint from $A + B$ and $A + C$ respectively, $B + C$ is disjoint from $B$ and $C$ respectively. Hence $B + C$ is contained in $\mathbb{Z}\setminus(B \cup C) = A$. Since $B + C = A + A$, we obtain $A + A \subset A$. On the other hand, $A = \{0\} + A \subset A + A$, so we obtain $A = A + A = B + C$.
Therefore $A+B+C = A+A+A = A$, and $B+B = C+A, C+C = A+B$
from which we can derive $B+B+B = A+B+C = A, C+C+C = A+B+C = A$.
In particular, if for any $r \in \mathbb{Q} = A \cup B \cup C$ then $3r \in A$.
But this conclusion is impossible. Take any $b \in B$ ($B$ being a nonempty subset) and let $r = \frac{b}{3} \in \mathbb{Q}$, then
$b = 3r \in A$ yields a contradiction.