6. (6 points) Define a new operation “*”: $a^{*} b=\left\{\begin{array}{l}\mathrm{a} \text { (if } \mathrm{a}>\mathrm{b}) \\ 1 \text { (if } \mathrm{a}=\mathrm{b}) \\ \mathrm{b} \text { (if } \mathrm{a}<\mathrm{b})\end{array}\right.$ For example, $3.5 * 2=3.5,1 * 1.2=1.2,7 * 7=1$, then $\frac{1.1 * \frac{7}{3}-\frac{1}{3} * 0.1}{\frac{4}{5} * 0.8}=$ $\qquad$
6. (6 points) Define a new operation “*”: $a^{*} b=\left\{\begin{array}{l}\mathrm{a}(\text { if } \mathrm{a}>\mathrm{b}) \\ 1 \text { (if } \mathrm{a}=\mathrm{b}) \\ \mathrm{b}(\text { if } \mathrm{a}<\mathrm{b})\end{array}\right.$
For example, $3.5 * 2=3.5,1 * 1.2=1.2,7 * 7=1$, then $\frac{1.1 * \frac{7}{3}-\frac{1}{3} * 0.1}{\frac{4}{5} * 0.8}=2$.
【Solution】Solution: According to the analysis, we have
$$
\begin{array}{l}
\frac{1.1 * \frac{7}{3}-\frac{1}{3} * 0.1}{\frac{4}{5} * 0.8}, \\
=\frac{\frac{7}{3}-\frac{1}{3}}{1}, \\
=2 ;
\end{array}
$$

The answer is: 2.