Example 11 A function $f(x)$ defined on $\mathbf{R}$ satisfies: for any real numbers $m, n$, it always holds that $f(m+n)=f(m) \cdot f(n)$, and when $x>0$, $0 < f(x) < 1$. Let $A=\{(x, y) ; y > \log_{f(1)} x\} \cdot B=\{(x, y) ; f(a x-y+\sqrt{2})=1, a \in \mathbf{R}\}$, if $A \cap B=\varnothing$, try to determine the range of values for $a$.
(1) In $f(m+n)=f(m) \cdot f(n)$, let $m=1, n=0$, we get $f(1)=f(1) \cdot f(0)$. Since $f(1) \neq 0$, it follows that $f(0)=1$.
In $f(m+n)=f(m) \cdot f(n)$, let $m=x, n=-x$.
Because when $x>0$, $00,01>0$.
Also, when $x=0$, $f(0)=1>0$, so, in summary, for any $x \in \mathbf{R}$, we have $f(x)>0$.
Let $-\infty0.0f(1)$
That is, $x^{2}+y^{2}<1$
Also, $f(a x-y+\sqrt{2})=1=f(0)$, according to the monotonicity of the function, we have $a x-y+\sqrt{2}=0$.
Since $A \cap B=\varnothing$, the line $a x-y+\sqrt{2}=0$ has no common points with the circle $x^{2}+y^{2}<1$. Therefore, $\frac{\sqrt{2}}{\sqrt{a^{2}}+1} \geqslant 1$, solving this gives $-3 \leqslant a \leqslant 3$.