5. Find all values of the parameter $a$ for which the equation $\left(\left(1-x^{2}\right)^{2}+2 a^{2}+5 a\right)^{7}-\left((3 a+2)\left(1-x^{2}\right)+3\right)^{7}=5-2 a-(3 a+2) x^{2}-2 a^{2}-\left(1-x^{2}\right)^{2}$ has two distinct solutions on the interval $\left[-\frac{\sqrt{6}}{2} ; \sqrt{2}\right]$. Specify these solutions for each found $a$.

$\quad \quad \quad$ Solution: $\quad\left(\left(1-x^{2}\right)^{2}+2 a^{2}+5 a\right)^{7}-\left((3 a+2)\left(1-x^{2}\right)+3\right)^{7}=5-2 a-(3 a+2) x^{2}-2 a^{2}-\left(1-x^{2}\right)^{2} \Leftrightarrow$ $\left(\left(1-x^{2}\right)^{2}+2 a^{2}+5 a\right)^{7}+\left(1-x^{2}\right)^{2}+2 a^{2}+5 a=\left((3 a+2)\left(1-x^{2}\right)+3\right)^{7}+(3 a+2)\left(1-x^{2}\right)+3$.

Consider the function $f(t)=t^{7}+t$. The function $f(t)$ is increasing on the entire number line.

Let $u=\left(1-x^{2}\right)^{2}+2 a^{2}+5 a, \quad v=(3 a+2)\left(1-x^{2}\right)+3$. Then we have the equation $f(u)=f(v)$, and due to the strict monotonicity of the function $f$, we arrive at the equation $u=v$, i.e.,
$\left(1-x^{2}\right)^{2}+2 a^{2}+5 a=(3 a+2)\left(1-x^{2}\right)+3 . \quad$ The last equation is equivalent to the following $\left(1-x^{2}\right)^{2}-(3 a+2)\left(1-x^{2}\right)+2 a^{2}+5 a-3=0$. It is necessary to find all values of the parameter $a$ for which this equation has two distinct solutions on the interval $\left[-\frac{\sqrt{6}}{2} ; \sqrt{2}\right]$. Let's make the substitution: $y=1-x^{2}$. We arrive at the equation $y^{2}-(3 a+2) y+2 a^{2}+5 a-3=0$. For the condition of the problem to be satisfied, one root of this equation must belong to the interval $[-0.5 ; 1)$, and the second, if it exists, must not belong to the interval $[-1 ; 1]$, or this equation must have two distinct solutions belonging to the set $[-1 ;-0.5) \cup\{1\}$. The discriminant of the equation $D=(3 a+2)^{2}-8 a^{2}-20 a+12=(a-4)^{2}$. The only solution will be at $a=4$, this solution is $y=7$. In all other cases, we have two distinct solutions $y_{1}=2 a-1$ and $y_{2}=a+3$.

1) $-0.5 \leq 2 a-11 \Rightarrow a \in[0.25 ; 1), 1-x^{2}=2 a-1, x_{1}=\sqrt{2-2 a}, x_{2}=-\sqrt{2-2 a}$.
2) $-0.5 \leq a+31 \Rightarrow a \in[-3.5 ;-2), \quad 1-x^{2}=a+3, x_{1}=\sqrt{-a-2}$, $x_{2}=-\sqrt{-a-2}$.

The equation cannot have two distinct solutions belonging to the set $[-1 ;-0.5) \cup\{1\}$.
Answer: $a \in[0.25 ; 1), x_{1}=\sqrt{2-2 a}, x_{2}=-\sqrt{2-2 a}$.

$$
a \in[-3.5 ;-2), x_{1}=\sqrt{-a-2}, x_{2}=-\sqrt{-a-2}
$$