$\begin{array}{l}\text { 8. Let } a_{k}=\frac{2^{k}}{3^{2^{k}}+1}(k \in \mathbf{N}) \text {, and let } \\ A=a_{0}+a_{1}+\cdots+a_{9}, B=a_{0} a_{1} \cdots a_{9} \text {. } \\ \text { Then } \frac{A}{B}=\end{array}$
8. $\frac{3^{20}-1}{2^{47}}-\frac{1}{2^{36}}$.

Notice that, for any number $m$ greater than 1, we have $\frac{1}{m-1}-\frac{1}{m+1}=\frac{2}{m^{2}-1}$.
Multiplying both sides of the above equation by $2^{k}$ and setting $m=3^{2^{k}}$, we get
$$
\frac{2^{k}}{3^{2^{k}}+1}=\frac{2^{k}}{3^{2^{k}}-1}-\frac{2^{k+1}}{3^{2^{k+1}}-1} \text {. }
$$

Substituting $k=0,1, \cdots, 9$ into equation (1) and then summing them up, we get
$$
A=\frac{1}{2}-\frac{2^{10}}{3^{2^{10}}-1} .
$$

Let $C=2^{0} \times 2^{1} \times \cdots \times 2^{9}=2^{45}$,
$$
\begin{array}{l}
D=\prod_{k=0}^{9}\left(3^{2^{k}}+1\right) \\
=\frac{1}{2}\left(3^{2^{0}}-1\right) \prod_{k=0}^{9}\left(3^{2^{k}}+1\right) \\
=\frac{1}{2}\left(3^{2^{1}}-1\right) \prod_{k=1}^{9}\left(3^{2^{k}}+1\right)=\cdots \\
=\frac{1}{2}\left(3^{2^{10}}-1\right) .
\end{array}
$$

Then $\frac{1}{B}=\frac{D}{C}=\frac{3^{2^{10}}-1}{2^{46}}$.
From equations (2) and (3), we get
$$
\begin{array}{l}
\frac{A}{B}=\left(\frac{1}{2}-\frac{2^{10}}{3^{2^{10}}-1}\right)\left(\frac{2^{2^{10}}-1}{2^{46}}\right) \\
=\frac{3^{2^{10}}-1}{2^{47}}-\frac{1}{2^{36}} .
\end{array}
$$