Find three different irrational numbers lying between $\sqrt{3}$ and $\sqrt{5}$.

An irrational number lying between $\sqrt{3}$ and $\sqrt{5}$

$=\sqrt{\sqrt{3} \cdot \sqrt{5}}$

$=\sqrt{\sqrt{15}}$

$=15^{\frac{1}{4}}$

An irrational number lying between $\sqrt{3}$ and $15^{\frac{1}{4}}$

$=\sqrt{\sqrt{3} \cdot 15^{\frac{1}{4}}}$

$=\sqrt{3^{\frac{1}{2}} \cdot 3^{\frac{1}{4}} \cdot 5^{\frac{1}{4}}}$

$=\sqrt{3^{\frac{3}{4}} \cdot 5^{\frac{1}{4}}}$

$=3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}$

An irrational number lying between $\sqrt{3}$ and $3^{\frac{3}{6}} \cdot 5^{\frac{1}{8}}$

$=\sqrt{\sqrt{3} \cdot 3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}}$

$=\sqrt{3^{\frac{1}{2}} \cdot 3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}}$

$=\sqrt{3^{\frac{7}{8}} \cdot 5^{\frac{1}{8}}}$

$=3^{\frac{7}{16}} \cdot 5^{\frac{1}{16}}$

Hence, required three irrational numbers between $\sqrt{3}$ and $\sqrt{5}$ are $15^{\frac{1}{4}}, 3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}$ and $3^{\frac{7}{16}} \cdot 5^{\frac{1}{16}}$