$A=\frac{3+\left( 2{{\cos }^{2}}2x-1 \right)}{2{{\sin }^{2}}2x}=\frac{2+2{{\cos }^{2}}2x}{2{{\sin }^{2}}2x}=\frac{1+{{\cos }^{2}}2x}{{{\sin }^{2}}2x}$  $=\frac{1}{{{\sin }^{2}}2x}+\frac{{{\cos }^{2}}2x}{{{\sin }^{2}}2x}=\left( 1+{{\cot }^{2}}2x \right)+{{\cot }^{2}}2x=1+2{{\cot }^{2}}2x$  $\tan 2x=\frac{2\tan x}{1-{{\tan }^{2}}x}=\frac{2\sqrt{2}}{1-{{\left( \sqrt{2} \right)}^{2}}}=\frac{2\sqrt{2}}{1-2}=-2\sqrt{2}$  $A=1+2{{\cot }^{2}}2x=1+\frac{2}{{{\tan }^{2}}2x}=1+\frac{2}{{{\left( -2\sqrt{2} \right)}^{2}}}=1+\frac{2}{8}=1+\frac{1}{4}=\frac{5}{4}$