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

To find an irrational number between two positive numbers $a$ and $b$,we can use the formula $\sqrt{a \cdot b}$.
$1$. First irrational number between $\sqrt{3}$ and $\sqrt{5}$:
$= \sqrt{\sqrt{3} \cdot \sqrt{5}} = \sqrt{\sqrt{15}} = 15^{\frac{1}{4}}$.
$2$. Second irrational number between $\sqrt{3}$ and $15^{\frac{1}{4}}$:
$= \sqrt{\sqrt{3} \cdot 15^{\frac{1}{4}}} = \sqrt{3^{\frac{1}{2}} \cdot 3^{\frac{1}{4}} \cdot 5^{\frac{1}{4}}} = \sqrt{3^{\frac{3}{4}} \cdot 5^{\frac{1}{4}}} = 3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}$.
$3$. Third irrational number between $\sqrt{3}$ and $3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}$:
$= \sqrt{\sqrt{3} \cdot 3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}} = \sqrt{3^{\frac{1}{2}} \cdot 3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}} = \sqrt{3^{\frac{7}{8}} \cdot 5^{\frac{1}{8}}} = 3^{\frac{7}{16}} \cdot 5^{\frac{1}{16}}$.
Thus,the three irrational numbers are $15^{\frac{1}{4}}$,$3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}$,and $3^{\frac{7}{16}} \cdot 5^{\frac{1}{16}}$.