8. Given $a, b \in R$. Try to find $s=\max \{|a+b|,|a-b|, |1-a|, |1-b|\}$, the minimum value.
If $a b>0$, then
$$
|a-b| \leqslant|a|+|b|=|a+b| \text {. }
$$

Thus, $s=\max \{|a+b|,|1-a|,|1-b|\}$
$$
\begin{array}{l}
>\frac{1}{3}[|a+b|+|1-a|+|1-b|] \\
>\frac{1}{3}|(a+b)+(1-a)+(1-b)| \\
=\frac{2}{3},
\end{array}
$$

Equality is achieved when $a=b=\frac{1}{3}$.
If $a b<0$, then $|a-b|>|a|+|b|$. Thus,
$$
S \geqslant \max \{|1-a|,|1-b|\}>1>\frac{2}{3} .
$$

In summary, the minimum value of $S$ is $\frac{2}{3}$.