1. $A, B, C, D, E$ five people participate in a chess tournament together and find that their average age is exactly 28 years old. One year later, $A, B, C, D, F$ participate together and find that their average age is exactly 30 years old. How many years older is $F$ than $E$?
II. 1.5 years.
Let the age of $E$ in the first year of the competition be $x$, and the sum of the ages of the other four people be $a$. In the second year of the competition, the age of $F$ is $y$. Then, from the given information, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{a+x}{5}=28, \\
\frac{a+4+y}{5}=30
\end{array} \Rightarrow y-x=6\right. \\
\Rightarrow y-(x+1)=5 .
\end{array}
$$

Therefore, $F$ is 5 years older than $E$.