$$A \equiv 3 \pmod{5}$$
$$B \equiv 4 \pmod{5}$$
- Determine the remainder of $A \cdot B$ and $A+B$.
Rewriting the equations
$$A-3 \equiv 0 \pmod{5}$$
$$B -4 \equiv 0 \pmod{5}$$
Which yields $A \in \{3,8\}$ and $B \in \{4,9\}$. As it appears, I've gone wrong somewhere due to that there are solutions more than one.