$\text{Đáp án :}$ $\\$ `1.` Ta có:`sin^2x+cos^2x=1` `=>cos^2x=1-sin^2x` `=>cos^2x=1-(1/2)^2` `=>cos^2x=3/4` Ta có:`1+tan^2x=1/(cos^2x)` `=>tan^2x=1/(cos^2x)-1` `=>tan^2x=`$\dfrac{1}{\dfrac{3}{4}}-1$ `=>tan^2x=4/3-1` `=>tan^2x=1/3` Ta có:`A=5.tan^2x-1/2 .cos^2x` `=5. 1/3-1/2 . 3/4` `=5/3-3/8` `=31/24` $\\$ `2.)` Ta có:`cosx=1/4` `=>cos^2x=(1/4)^2` `=>cos^2x=1/16` Ta có:`sin^2x+cos^2x=1` `=>sin^2x=1-cos^2x` `=>sin^2x=1-1/16` `=>sin^2x=15/16` Ta có:`A=3.cos^2x-5/2 .sin^2x` `=3. 1/16-5/2 . 15/16` `=3/16-75/32` `=-69/32` $\\$ $\textit{#1412}$