Do the 12 question.

Dear student,
Here is the solution of your asked query:
$$\begin{gathered} \mathrm{Since} \sqrt{5} \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as} 5^{\frac{1}{2}}, \mathrm{so} \mathrm{we} \mathrm{have}; \\ \frac{\sqrt{5\sqrt{5\sqrt{5\sqrt{5\sqrt{5}}}}}}{\sqrt{5\sqrt{5\sqrt{5}}}} \\ = \frac{\sqrt{5\sqrt{5\sqrt{5\sqrt{5\times 5^{{\displaystyle \frac{1}{2}}}}}}}}{\sqrt{5\sqrt{5\times 5^{{\displaystyle \frac{1}{2}}}}}} \\ = \frac{\sqrt{5\sqrt{5\sqrt{5\sqrt{5^{{\displaystyle 1+\frac{1}{2}}}}}}}}{\sqrt{5\sqrt{5^{{\displaystyle 1+\frac{1}{2}}}}}} \{\mathrm{using} {\mathrm{a}}^{\mathrm{m}}{\times }^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\} \\ = \frac{\sqrt{5\sqrt{5\sqrt{5\times {\left(5^{{\displaystyle \frac{3}{2}}}\right)}^{{\displaystyle \frac{1}{2}}}}}}}{\sqrt{5\times {\left(5^{\frac{3}{2}}\right)}^{\frac{1}{2}}}} \{\mathrm{using} {\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}\times \mathrm{n}}\} \\ = \frac{\sqrt{5\sqrt{5\sqrt{5\times 5^{{\displaystyle \frac{3}{4}}}}}}}{\sqrt{5\times 5^{{\displaystyle \frac{3}{4}}}}} \\ = \frac{\sqrt{5\sqrt{5\times {\left(5^{1+{\displaystyle \frac{3}{4}}}\right)}^{{\displaystyle \frac{1}{2}}}}}}{{\left(5^{1+\frac{3}{4}}\right)}^{\frac{1}{2}}} \\ = \frac{\sqrt{5\sqrt{5\times 5^{{\displaystyle \frac{7}{8}}}}}}{5^{{\displaystyle \frac{7}{8}}}} \\ = \frac{\sqrt{5\times {\left(5^{1+{\displaystyle \frac{7}{8}}}\right)}^{{\displaystyle \frac{1}{2}}}}}{5^{{\displaystyle \frac{7}{8}}}} \\ = \frac{\sqrt{5\times 5^{{\displaystyle \frac{15}{16}}}}}{5^{{\displaystyle \frac{7}{8}}}} \\ = \frac{{\left(5^{1+{\displaystyle \frac{15}{16}}}\right)}^{{\displaystyle \frac{1}{2}}}}{5^{{\displaystyle \frac{7}{8}}}} \\ = \frac{5^{{\displaystyle \frac{31}{32}}}}{5^{{\displaystyle \frac{7}{8}}}} \\ = 5^{\frac{31}{32}-\frac{7}{8} } \{\mathrm{using} \frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\} \\ = 5^{\frac{3}{32}} \end{gathered}$$
Regards