8. DOD: $a>0 ; a \neq 1 ; x+a>0$.

From uniqueness, it follows that $y=0$, so we get $\left\{\begin{array}{l}\log _{2}(x+a)=2 \\ |x|=2\end{array} \Rightarrow\left\{\begin{array}{l}x+a=4 \\ x= \pm 2\end{array} \Rightarrow \boldsymbol{I}\left\{\begin{array}{l}x=2 \\ \boldsymbol{a}=2\end{array} \quad \boldsymbol{I}\left\{\begin{array}{l}x=-2 \\ \boldsymbol{a}=6\end{array}\right.\right.\right.\right.$

Let's check by direct substitution into the system

For $a=2\left\{\begin{array}{l}\log _{2}(x+2)+2|y|=2 \\ \log _{2}\left(y^{2}+1\right)+|x|=2\end{array}\right.$ the system will have at least two solutions $(2 ; 0) u(-1 ; 1)$. Therefore, $a=2$ does not work.

For $a=6\left\{\begin{array}{l}\log _{2}(x+6)+2|y|=2 \\ \log _{6}\left(y^{2}+1\right)+|x|=2\end{array} \Rightarrow\left\{\begin{array}{l}\log _{2}(x+6)=2(1-|y|) \\ \log _{6}\left(y^{2}+1\right)=2-|x|\end{array} \Rightarrow\right.\right.$

Since $\log _{6}\left(y^{2}+1\right) \geq 0$, then $|x| \leq 2$. Therefore, $2 \leq \log _{2}(\boldsymbol{x}+6) \leq 3 \Rightarrow 2 \leq 2(1-|\boldsymbol{y}|) \leq 3 \Rightarrow|\boldsymbol{y}| \leq 0 \Rightarrow \boldsymbol{y}=0 ; \boldsymbol{x}=-2$.

Thus, for $a=6$ we have the unique solution $x=-2, y=0$.
Answer: $a=6, x=-2, y=0$.

## 3rd variant.