11 Seven positive numbers form a geometric sequence $a_{1}, a_{2}, \cdots, a_{7}$, the sum of the first five terms is $\frac{62}{7 \sqrt{2}-6}$, the sum of the last five terms is $12+14 \sqrt{2}$, then the product of these seven terms $a_{1} a_{2} \cdots a_{7}=$ $\qquad$
11 Let the common ratio be $q$, then we have
$$
a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=q^{2}\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right),
$$

which means
$$
12+14 \sqrt{2}=\frac{62}{7 \sqrt{2}-6} \cdot q^{2},
$$

from which we get $q=\sqrt{2}$. From
$$
q^{2}\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)=12+14 \sqrt{2},
$$

we get $a_{1}=\sqrt{2}$, hence the terms are $(\sqrt{2})^{k}(k=1,2, \cdots, 7)$, and their product is $2^{14}$.