5. The first equation is reduced to the form $(x+y)^{2}+(y-a)^{2}=0$, from which we obtain $x=-a ; y=a$. Substitution into the inequality gives:

$$
2^{-2-a} \cdot \log _{2}(-a)<1 \Longleftrightarrow \log _{2}(-a)<2^{2+a}
$$

Equality is achieved at $a=-2$, and due to monotonicity we get $a \in$ $(-2 ; 0)$.
Answer: $x=-a ; y=a$ for $a \in(-2 ; 0)$.

Answer to variant: $4-2: x=a ; y=-a$ for $a \in(-3 ; 0)$.