8. DOD: $a>0 ; \quad a \neq 1 ; \quad 4 y+a^{2}>0$.

Let $\left(x_{0} ; y_{0}\right)$ be a solution to the given system. Then, due to the evenness of the system with respect to $x$, $\left(-x_{0} ; y_{0}\right)$ will also be a solution. Therefore, the necessary condition for uniqueness is $x_{0}=0$. Substituting $x_{0}=0$ into the second equation, we get $y= \pm 1$. By substituting the pairs $(0 ; 1)$ and $(0 ;-1)$ into the first equation of the system, we will find for which $a$ they are solutions of the system.

For the pair $(0 ; 1)$, we find $\mathrm{a}= \pm 2$, and for the pair $(0 ;-1)$, we find that $\boldsymbol{a}= \pm 2 \sqrt{3}$. The values $a=-2$ and $a=-2 \sqrt{3}$ contradict the definition of the logarithm, so they are not the solutions we are looking for. For $a=2$, we have $\left\{\begin{array}{l}|x|+\log _{2}(4 y+4)=3 \\ \log _{2}\left(x^{2}+1\right)+y^{2}=1 .\end{array}\right.$ It is easy to see that this system has at least two solutions. One is $(0 ; 1)$, the other is $(1 ; 0)$, i.e., $a=2$ is not the solution we are looking for.

For $a=2 \sqrt{3}$, we get: $\left\{\begin{array}{l}|x|+\log _{2}(4 y+12)=3 \\ \log _{2 \sqrt{3}}\left(x^{2}+1\right)+y^{2}=1 .\end{array}\right.$

On $\mathrm{R}$, the function $\log _{2 \sqrt{3}}\left(\boldsymbol{x}^{2}+1\right) \geq 0 \Rightarrow-1 \leq \boldsymbol{y} \leq 1 \Rightarrow 3 \leq \log _{2}(4 \boldsymbol{y}+12) \leq 4 \Rightarrow$

Then the first equation has a solution $x=0$, so $\log _{2}(4 \boldsymbol{y}+12)=3 \Rightarrow \boldsymbol{y}=-1$. Therefore, the system has a unique solution when $\boldsymbol{a}=2 \sqrt{3}$, and this solution is the pair $(0 ;-1)$.
Answer: $\boldsymbol{a}=2 \sqrt{3}, x=0 ; y=-1$.

2nd variant.