8. Given $a \geqslant b \geqslant c \geqslant d \geqslant 0$,
$$
\frac{a^{2}+b^{2}+c^{2}+d^{2}}{(a+b+c+d)^{2}}=\frac{3}{8} \text {. }
$$

Then the maximum value of $\frac{a+c}{b+d}$ is
8. 3 .

If $c, d$ are zero, it does not meet the conditions, so only $d=0$.
Let $x=a+c, y=b+d$.
Since $a \geqslant b \geqslant c \geqslant d \geqslant 0$, then,
$(a-d)(b-c) \geqslant 0$
$\Leftrightarrow 2(a c+b d) \leqslant(a+d)(b+c)$.
By the mean value inequality, we get
$$
\begin{array}{l}
2(a c+b d) \leqslant(a+d)(b+c) \\
\leqslant \frac{1}{4}(a+b+c+d)^{2}=\frac{1}{4}(x+y)^{2} .
\end{array}
$$

Then, by the given conditions, we have
$$
\begin{array}{l}
(x+y)^{2}=(a+b+c+d)^{2} \\
=\frac{8}{3}\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \\
=\frac{8}{3}\left(x^{2}+y^{2}\right)-\frac{16}{3}(a c+b d) \\
\quad \geqslant \frac{8}{3}\left(x^{2}+y^{2}\right)-\frac{2}{3}(x+y)^{2} .
\end{array}
$$

Let $t=\frac{x}{y}=\frac{a+c}{b+d}$. Then
$$
\begin{array}{l}
5(x+y)^{2} \geqslant 8\left(x^{2}+y^{2}\right) \\
\Leftrightarrow 3 t^{2}-10 t+3 \leqslant 0 \Rightarrow t \leqslant 3 .
\end{array}
$$

When $a=3, d=0, b=c=\frac{3}{2}$, the maximum value of $\frac{a+c}{b+d}$ is 3.