Помогите пожалуйста

Решите задачу:

$========= 1 =========\\\\ 0.5 \sqrt{0.04} + \frac{1}{6} \sqrt{144} = 0.5 \sqrt{0.2^2} + \frac{1}{6} \sqrt{12^2} =0.5\cdot0.2 + \frac{1}{6} \cdot 12=\\ =0.1+2=2.1\\\\ 2 \sqrt{1 \frac{9}{16} } -1=2 \sqrt{\frac{25}{16} } -1=2 \sqrt{(\frac{5}{4})^2 } -1=2\cdot \frac{5}{4}-1= \frac{5}{2}-1= \frac{3}{2}=1.5\\\\ (2 \sqrt{0.5} )^2=2^2\cdot (\sqrt{0.5} )^2=4\cdot 0.5=2\\\\$
$========= 2 =========\\\\ \sqrt{0.25\cdot64} = \sqrt{0.5^2\cdot8^2} = 0.5\cdot8=4\\\\ \sqrt{56} \cdot \sqrt{14} = \sqrt{56 \cdot 14} = \sqrt{784}= \sqrt{28^2} =28\\\\ \sqrt{3^4 \cdot 2^6} = \sqrt{3^4 \cdot 2^4 \cdot 2^2} = \sqrt{6^4 \cdot 2^2} = 6^2\cdot 2=36 \cdot 2 = 72\\\\ \frac{ \sqrt{8} }{ \sqrt{2} } = \sqrt{ \frac{8}{2} } = \sqrt{4} = \sqrt{2^2}=2\\\\$
$========= 3 =========\\\\ x^2=49\\ x=\pm \sqrt{49} \\ x=\pm \sqrt{7^2} \\ x=\pm 7 \rightarrow x_1=7, x_2=-7\\\\ x^2=10\\ x=\pm \sqrt{10} \rightarrow x_1= \sqrt{10} , x_2=- \sqrt{10} \\\\$
$========= 4 =========\\\\ x^2 \sqrt{9x^2}= x^2 \sqrt{(3x)^2}= x^2\cdot(3x)=3x^3\\\\ -5b^2 \sqrt{ \frac{4}{b^2} } =-5b^2 \sqrt{ (\frac{2}{-b})^2 } =-5b^2\cdot(\frac{2}{-b})=10b$