5. Solution: Consider the second equation of the system $3-x=\sqrt{x^{2}-6 x+y+8} \Leftrightarrow\left\{\begin{array}{l}x \leq 3, \\ y=1 .\end{array}\right.$

Therefore, $\left\{\begin{array}{c}\log _{|x-1|}(a x)=2 \log _{|x-1|}(x+y), \\ 3-x=\sqrt{x^{2}-6 x+y+8}\end{array} \Leftrightarrow\left\{\begin{array}{c}x^{2}+(2-a) x+1=0, \\ y=1, \\ -1<x \leq 3, x \neq 0, x \neq 1, x \neq 2 .\end{array}\right.\right.$

Determine for which values of the parameter $a$ the equation $x^{2}+(2-a) x+1=0(*)$ has a unique solution if $-1<x \leq 3, x \neq 0, x \neq 1, x \neq 2$.

1) $D=a(a-4), D=a(a-4), x=\frac{a-2}{2}$. For $a=0$ the root $x=-1$ does not fit; for $a=4$ the root $x=1$ does not fit.
2) Determine for which $a$ the points $x=0, x=1, x=2$ are solutions of equation (*). $x=0$ is not a solution for any $a$; $x=1$ is the only solution of equation (*) for $a=4$; $x=2$ is a solution of (*) for $a=4.5$, since substituting $x=2$ into equation $\left.{ }^{*}\right)$ gives $2^{2}+(2-a) 2+1=0,2 a=9$. However, for $a=4.5$ equation (*) has a second solution $x=0.5$, which satisfies the given conditions.

Therefore, for $a=4.5$ the system has a unique solution $x=0.5, y=1$.

3) If the discriminant of equation (*) is greater than zero, then the equation has two distinct solutions, but under the condition $f(-1) f(3)<0$, where $f(x)=x^{2}+(2-a) x+1$, one root will be extraneous, and one will satisfy the inequality $-1<x<3$. We have $f(-1)=a$, $f(3)=16-3 a$, leading to the inequality $a(16-3 a)<0$, and $a \in(-\infty ; 0) \cup(16 / 3 ;+\infty)$. If $a \in(-\infty ; 0)$, then $x=\frac{a-2+\sqrt{a^{2}-4 a}}{2}$.

If $a \in(16 / 3 ;+\infty)$, then $x=\frac{a-2-\sqrt{a^{2}-4 a}}{2}$.

4) Check the cases when $f(-1)=0$ and $f(3)=0$. The first equality holds for $a=0$, in which case equation (*) has no solutions that satisfy the given conditions. The second equality is true for $a=16 / 3$. In this case, equation (*) becomes $x^{2}+\frac{10}{3} x+1=0$, and has two solutions $x=1 / 3$ and $x=3$, both of which fit.
Answer: for $a=4.5 \quad x=0.5, y=1$, for $a \in(-\infty ; 0) \quad x=\frac{a-2+\sqrt{a^{2}-4 a}}{2}, y=1$, for $a \in(16 / 3 ;+\infty) \quad x=\frac{a-2-\sqrt{a^{2}-4 a}}{2}, y=1$.