(9) Given the function $f(x)=x^{2}-a x+a-1, a \in \mathbf{R}$, if for any $a \in(0,4)$, there exists $x_{0} \in[0,2]$, such that $t \leqslant\left|f\left(x_{0}\right)\right|$ holds, then the range of values for $t$ is . $\qquad$
Solution: $t \leqslant 1$.

Solution 1: The function $f(x)$ is regarded as a function of $a$, $f(a)=(1-x)[a+(x+1)]$.
The problem is equivalent to $\forall a \in(0,4], t \leqslant|f(a)|$.
Since $x \in[0,2]$, $|f(a)|_{\max }=\left|x^{2}-4 x+3\right|$.
Therefore, the problem is equivalent to $\exists x \in[0,2], t \leqslant|f(a)|_{\max }=\left|x^{2}-4 x+3\right|$.
That is, $t \leqslant\left|x^{2}-4 x+3\right|_{\min }$, so $t \leqslant 1$.

Solution 2: From the problem, we have $f(x)=(x+a-1)(x-1)$.
For $\exists x_{0} \in[0,2]$, it is only necessary that $t \leqslant \max \left\{|a-1|,|3-a|,\left|-\frac{1}{4} a^{2}+a-1\right|\right\}$.
Let $g(a)=\max \left\{|a-1|,|3-a|,\left|-\frac{1}{4} a^{2}+a-1\right|\right\}$.
At this time, $g(a)_{\min }=1$.
Therefore, for $\forall a \in(0,4)$, it is only necessary that $t \leqslant g(a)_{\min }$, so $t \leqslant 1$.

Solution 3: If for any $a \in(0,4)$, there exists $x_{0} \in[0,2]$ such that $t \leqslant\left|f\left(x_{0}\right)\right|$.
When $t \leqslant 0$, the condition is satisfied;
When $t>0$, it is equivalent to if for any $a \in(0,4)$, there exists $x_{0} \in[0,2]$ such that $t \leqslant f\left(x_{0}\right)$ or $f\left(x_{0}\right) \leqslant-t$.
If $t \leqslant f\left(x_{0}\right)$, then because $a \in(0,4)$, $\frac{a}{2} \in(0,2)$.
Therefore, the function $f(x)=x^{2}-a x+a-1$ is decreasing in $\left(0, \frac{a}{2}\right)$ and increasing in $\left(\frac{a}{2}, 2\right)$.

So when $a \in(0,2]$, $t \leqslant 3-a$ and thus $00$ is a contradiction, discard it.
In summary: $t \leqslant 1$.