12. Let $a \in \mathbf{R}$, and for any real number $b$ we have $\max _{x \in[0,1]}\left|x^{2}+a x+b\right| \geq 1$, find the range of values for $a$.
Solution 1 Let $f(x)=x^{2}+a x+b$, for $|b| \geq 1 \Rightarrow|f(0)| \geq 1$, so we only need to consider $|b|f(0)=b$, so $f(1)=1+a+b \geq 1$, solving gives $a \geq 1$;
(2) When $0<-\frac{a}{2} \leq \frac{1}{2}$, i.e., $-1 \leq a<0$, at this time the maximum and minimum values of the function $f(x)$ are obtained at the vertex and the right endpoint of the parabola, and for $b=0$ we have $|f(1)|=|1+a|<1, \quad\left|f\left(-\frac{a}{2}\right)\right|=\left|-\frac{a^{2}}{4}\right|<1$;
(3) When $\frac{1}{2}<-\frac{a}{2} \leq 1$, i.e., $-2 \leq a<-1$, at this time the maximum and minimum values of the function $f(x)$ are obtained at the vertex and the left endpoint of the parabola, and for $b=0$ we have $|f(0)|=|b|<1,\left|f\left(-\frac{a}{2}\right)\right|=\left|-\frac{a^{2}}{4}\right|<1$;
(4) When $-\frac{a}{2} \geq 1$, i.e., $a \leq-2$, at this time the maximum and minimum values of the function $f(x)$ are obtained at the left and right endpoints of the parabola, for any $|b|<1$ we have $|f(0)|=|b|<1$, so $f(1)=1+a+b \leq-1$, solving gives $a \leq-3$.
In summary, $a \geq 1$ or $a \leq-3$.
(20 points)
Solution 2 Let $m=\max _{x \in[0,1]}\left|x^{2}+a x+b\right|$, then we have
$$
m \geq|b|, m \geq|1+a+b| \Rightarrow 2 m \geq|b|+|1+a+b| \geq|1+a|
$$

According to the problem, $\frac{|1+a|}{2} \geq 1 \Rightarrow a \geq 1$ or $a \leq-3$.