Answer: Option A. -> 0
$$\eqalign{
& = \sqrt {0.01} \times \root 3 \of {0.008} - 0.02{\text{ }} \cr
& = \sqrt {{{\left( {0.1} \right)}^2}} \times \root 3 \of {{{\left( {0.2} \right)}^3}} - 0.02 \cr
& = 0.1 \times 0.2 - 0.02 \cr
& = 0.02 - 0.02 \cr
& = 0 \cr} $$

Answer: Option B. -> 0.2
$$\eqalign{
& = \sqrt {0.000064} \cr
& = \sqrt {\frac{{64}}{{{{10}^6}}}} \cr
& = \frac{8}{{{{10}^3}}} \cr
& = \frac{8}{{1000}} \cr
& = 0.008 \cr
& \therefore \root 3 \of {\sqrt {0.000064} } \cr
& = \root 3 \of {0.008} \cr
& = \root 3 \of {\frac{8}{{1000}}} \cr
& = \frac{2}{{10}} \cr
& = 0.2 \cr} $$

Answer: Option C. -> 0.5
$$\eqalign{
& = \root 3 \of {\frac{{0.2 \times 0.2 \times 0.2 + 0.04 \times 0.04 \times 0.04}}{{0.4 \times 0.4 \times 0.4 + 0.08 \times 0.08 \times 0.08}}} \cr
& = \root 3 \of {\frac{{0.008 + 0.000064}}{{0.064 + 0.000512}}} \cr
& = \root 3 \of {\frac{{0.008064}}{{0.064512}}} \cr
& = \root 3 \of {\frac{{8064}}{{64512}}} \cr
& = \root 3 \of {\frac{1}{8}} \cr
& = \frac{1}{2} \cr
& = 0.5 \cr} $$

Answer: Option D. -> 450
$$3600 = {2^3} \times {5^2} \times {3^2} \times 2$$
To make it a perfect cube, it must be divided by   $${5^2} \times {3^2} \times 2,i.e.,450$$

Answer: Option C. -> $${10^{ - 3}}$$
$$\eqalign{
& \Leftrightarrow 37 + \sqrt {0.0615 + x} = 37.25 \cr
& \Leftrightarrow \sqrt {0.0615 + x} = 0.25 \cr
& \Leftrightarrow 0.0615 + x = {\left( {0.25} \right)^2} = 0.0625 \cr
& \Leftrightarrow x = 0.001 \cr
& \Leftrightarrow x = \frac{1}{{{{10}^3}}} \cr
& \Leftrightarrow x = {10^{ - 3}} \cr} $$

Answer: Option A. -> 0.49
$$\eqalign{
& {\text{Let ,}} \cr
& \sqrt {\frac{{0.0196}}{x}} = 0.2 \cr
& {\text{Then,}} \cr
& \Leftrightarrow \frac{{0.0196}}{x} = 0.04 \cr
& \Leftrightarrow x = \frac{{0.0196}}{{0.04}} \cr
& \Leftrightarrow x = \frac{{1.96}}{4} \cr
& \Leftrightarrow x = 0.49 \cr} $$

Answer: Option A. -> 14.5
$$\eqalign{
& {\text{Let the number be }}x \cr
& {\text{Then,}} \cr
& \Leftrightarrow \frac{3}{5}{x^2} = 126.15 \cr
& \Leftrightarrow {x^2} = \left( {126.15 \times \frac{5}{3}} \right) \cr
& \Leftrightarrow {x^2} = 210.25 \cr
& \Leftrightarrow x = \sqrt {210.25} \cr
& \Leftrightarrow x = 14.5 \cr} $$

Answer: Option C. -> 27
$$\eqalign{
& \Rightarrow \sqrt {1 + \frac{x}{{169}}} = \frac{{14}}{{13}} \cr
& \Rightarrow 1 + \frac{x}{{169}} = \frac{{196}}{{169}} \cr
& \Rightarrow \frac{x}{{169}} = \left( {\frac{{196}}{{169}} - 1} \right) \cr
& \Rightarrow \frac{x}{{169}} = \frac{{27}}{{169}} \cr
& \Rightarrow x = 27 \cr} $$

Answer: Option A. -> 8, 5
$$\eqalign{
& \Leftrightarrow \sqrt {\left( {x - 1} \right)\left( {y + 2} \right)} = 7 \cr
& \Leftrightarrow \left( {x - 1} \right)\left( {y + 2} \right) = {\left( 7 \right)^2} \cr
& \Leftrightarrow \left( {x - 1} \right) = 7{\text{ and}}\left( {y + 2} \right) = 7 \cr
& \Leftrightarrow x = 8{\text{ and }}y = 5 \cr} $$

Answer: Option D. -> 1.414
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,1|\overline 2 \,.\,\,\overline {00} \,\,\overline {00} \,\,\overline {00} \,(1.414 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,|1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& \,\,\,\,\,\,24|\,\,1\,00 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,96 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr
& \,\,\,281\,|\,\,\,\,\,\,\,\,\,400 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,281 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr
& 2824\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11900 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11296 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr
& \therefore \sqrt 2 = 1.414 \cr} $$