1. Ellipse eccentricity

Given ellipse: $\frac{x^2}{25}+\frac{y^2}{b^2}=1, \quad b<5$

Here the semi-major axis is $a=5, \quad \text{and} \quad b<5$ so its eccentricity is $e_1=\frac{c_1}{a}=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{25-b^2}}{5}.$

2. Hyperbola eccentricity

Given hyperbola: $\frac{x^2}{16}-\frac{y^2}{b^2}=1$

For this hyperbola, $a=4$ and $c_2=\sqrt{a^2+b^2}=\sqrt{16+b^2}.$ So its eccentricity is $e_2=\frac{c_2}{a}=\frac{\sqrt{16+b^2}}{4}.$

3. Use the condition $e_1e_2=1$

We have $\frac{\sqrt{25-b^2}}{5}\cdot \frac{\sqrt{16+b^2}}{4}=1.$

So, $\sqrt{(25-b^2)(16+b^2)}=20.$

Squaring both sides, $(25-b^2)(16+b^2)=400.$

Expand: $400+25b^2-16b^2-b^4=400$ $400+9b^2-b^4=400$ $9b^2-b^4=0$ $b^2(9-b^2)=0.$

Thus, $b^2=0 \quad \text{or} \quad b^2=9.$

Since $b$ is a positive semi-axis length, we take $b=3.$

4. Distance between foci of the ellipse

For the ellipse, $c_1=\sqrt{25-9}=4.$ Distance between foci is $\alpha=2c_1=8.$

5. Distance between foci of the hyperbola

For the hyperbola, $c_2=\sqrt{16+9}=5.$ Distance between foci is $\beta=2c_2=10.$

6. Final ordered pair

Therefore, $(\alpha,\beta)=(8,10).$

So the correct option is A.