8. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=$ $5, A D=3, A A_{1}=4$, a plane $\alpha$ passing through point $A$ and parallel to line $C D$ divides the rectangular prism into two parts. Now, two spheres are placed into these two geometric bodies, respectively. Then, during the variation of plane $\alpha$, the maximum value of the sum of the radii of these two spheres is $\qquad$
8. $\frac{21}{10}$.

Figure 2 is a cross-sectional diagram through the centers of two spheres and parallel to the left and right faces of the cuboid (with edge lengths 3 and 4).
Let $\angle O_{2} M E=\theta$.
Then $\angle M F H=2 \theta$.
Let $H F=x(0 < x \leq \frac{3}{5})$.
$$
f(x)=\frac{3}{5} \cdot \frac{1}{\sqrt{1-x^{2}}}
$$
Since $0 < x \leq \frac{3}{5}$, we have $0 < x^{2} \leq \frac{9}{25}$, and thus $1-x^{2} \geq \frac{16}{25}$.
Therefore, $\sqrt{1-x^{2}} \geq \frac{4}{5}$, and $\frac{1}{\sqrt{1-x^{2}}} \leq \frac{5}{4}$.
Hence, $f(x) \leq \frac{3}{5} \cdot \frac{5}{4} = \frac{3}{4} < \frac{21}{10}$.
Let $t=\frac{1}{\sqrt{1-x^{2}}}$, then $t \in \left[ \frac{5}{4}, +\infty \right)$.
Thus, $f(t)=\frac{3}{5} t$.
Since $f'(t)=\frac{3}{5} > 0$, $f(t)$ is an increasing function on the interval $\left(0, \frac{3}{5}\right]$.
Therefore, when $t=\frac{3}{5}$, $f(t)$ reaches its maximum value $f\left(\frac{3}{5}\right)=\frac{21}{10}$.