The uploaded image contains a series of mathematical problems involving simplifications, root expressions, and square root operations. Let's work through them systematically. I'll start with the first section and break down the problems into solutions.

1. Simplify:

a. $\sqrt{7} \times \sqrt{7}$

$\sqrt{7} \times \sqrt{7} = \sqrt{7^2} = 7$

b. $4 \sqrt{5} \times \sqrt{5}$

$4 \sqrt{5} \times \sqrt{5} = 4 \times 5 = 20$

c. $(3 \sqrt{3})^2$

$(3 \sqrt{3})^2 = 3^2 \times (\sqrt{3})^2 = 9 \times 3 = 27$

d. $(\sqrt{6})^4$

$(\sqrt{6})^4 = (\sqrt{6}^2)^2 = 6^2 = 36$

e. $(\sqrt{2})^5$

$(\sqrt{2})^5 = (\sqrt{2}^4) \cdot \sqrt{2} = 4 \cdot \sqrt{2} = 4\sqrt{2}$

f. $(2 \sqrt{3})^3$

$(2 \sqrt{3})^3 = 2^3 \cdot (\sqrt{3})^3 = 8 \cdot 3\sqrt{3} = 24\sqrt{3}$

g. $\sqrt{2} \times \sqrt{8}$

$\sqrt{2} \times \sqrt{8} = \sqrt{2 \cdot 8} = \sqrt{16} = 4$

h. $2 \sqrt{3} \times \sqrt{27}$

$2 \sqrt{3} \times \sqrt{27} = 2 \cdot \sqrt{3 \cdot 27} = 2 \cdot \sqrt{81} = 2 \cdot 9 = 18$

i. $\frac{\sqrt{32}}{\sqrt{2}}$

$\frac{\sqrt{32}}{\sqrt{2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$

j. $\frac{\sqrt{3}}{\sqrt{12}}$

$\frac{\sqrt{3}}{\sqrt{12}} = \sqrt{\frac{3}{12}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

k. $(\sqrt{6})^3$

$(\sqrt{6})^3 = (\sqrt{6}^2) \cdot \sqrt{6} = 6 \cdot \sqrt{6} = 6\sqrt{6}$

l. $(3 \sqrt{2})^3$

$(3 \sqrt{2})^3 = 3^3 \cdot (\sqrt{2})^3 = 27 \cdot 2\sqrt{2} = 54\sqrt{2}$

Would you like me to continue solving the remaining sections (2, 3, and 4), or focus on specific problems? Let me know if you'd like explanations for any particular steps!

Related Questions:

1. How do you simplify products of square roots, like $\sqrt{a} \times \sqrt{b}$?
2. What are the rules for exponents when applied to square roots?
3. How do you handle fractions involving square roots?
4. Can you explain how to express a square root in the form $k\sqrt{2}$?
5. Why is $(\sqrt{a})^n = a^{n/2}$ important in solving these problems?

Tip: Always check for perfect squares within square roots to simplify expressions efficiently!