Answer: B.
Solution: $A F=D E=b, E F=A F-A E=b-a$,
Since
$$
S_{\text {square } A B C D}=13 S_{\text {square } E F G H},
$$

That is, $S_{\text {square } A B C D}=4 S_{\triangle A D E}+S_{\text {square } E F G H}=13 S_{\text {square } E F G H}$,
Therefore,
$$
S_{\triangle A D E}=3 S_{\text {square } E F G H},
$$

That is,
$$
\begin{array}{c}
\frac{1}{2} a b=3(b-a)^{2}=3 b^{2}-6 a b+3 a^{2}, \\
a b=6 b^{2}-12 a b+6 a^{2}, \\
6 a^{2}-13 a b+6 b^{2}=0, \\
(2 a-3 b)(3 a-2 b)=0,
\end{array}
$$

Since
$$
0<a<b,
$$

Therefore,
$$
2 a<3 b, 2 a-3 b \neq 0,
$$

Thus,
$$
3 a-2 b=0,
$$

That is,
$$
\frac{a}{b}=\frac{2}{3} .
$$

Hence, the answer is B.
29. As shown in the figure, it is a diagram made by ancient Chinese mathematicians when proving the Pythagorean theorem. The four right-angled triangles in the figure are congruent. If the area of the larger square $ABCD$ is 13 times the area of the smaller square $EFGH$, and if $AE=a, DE=b$, then $\frac{a}{b}=(\quad)$
(A). $\frac{1}{2}$
(B). $\frac{2}{3}$
(C). $\frac{2}{5}$
(D). $\frac{3}{5}$