11. Given $x=\frac{b}{a}, a, b$ are coprime positive integers, and $a \leqslant 8, \sqrt{2}-1<x<\sqrt{3}-1$.
(1) Write down one value of $x$ that satisfies the condition;
(2) Find all values of $x$ that satisfy the condition.
3.11. (1) $x=\frac{1}{2}$ satisfies the condition.
(2) Since $x=\frac{b}{a}(a, b$ are coprime positive integers, and $a \leqslant$ $8)$, therefore,
$$
(\sqrt{2}-1) a<b<(\sqrt{3}-1) a .
$$

When $a=1$, $(\sqrt{2}-1) \times 1<b<(\sqrt{3}-1) \times 1$, there is no such positive integer $b$;

When $a=2$, $(\sqrt{2}-1) \times 2<b<(\sqrt{3}-1) \times 2$, hence $b$ $=1$, at this time, $x=\frac{1}{2}$;

When $a=3$, $(\sqrt{2}-1) \times 3<b<(\sqrt{3}-1) \times 3$, hence $b$ $=2$, at this time, $x=\frac{2}{3}$;

When $a=4$, $(\sqrt{2}-1) \times 4<b<(\sqrt{3}-1) \times 4$, there is no coprime positive integer $b$ with $a$;

When $a=5$, $(\sqrt{2}-1) \times 5<b<(\sqrt{3}-1) \times 5$, hence $b$ $=3$, at this time, $x=\frac{3}{5}$;

When $a=6$, $(\sqrt{2}-1) \times 6<b<(\sqrt{3}-1) \times 6$, there is no coprime positive integer $b$ with $a$;

When $a=7$, $(\sqrt{2}-1) \times 7<b<(\sqrt{3}-1) \times 7$, hence $b$ $=3,4,5$, at this time, $x=\frac{3}{7}, \frac{4}{7}, \frac{5}{7}$;

When $a=8$, $(\sqrt{2}-1) \times 8<b<(\sqrt{3}-1) \times 8$, hence $b$ $=5$, at this time, $x=\frac{5}{8}$.
Therefore, all fractions that satisfy the condition are
$$
\frac{1}{2}, \frac{2}{3}, \frac{3}{5}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{5}{8} \text {. }
$$