$a)$  

$S+T=\left(\frac{1}{2}{x}^3-2x^2+3x\right)+\left(\frac{1}{4}{x}^3-\frac{3}{2}{x}^2-6x+8\right)=$  

$=\underline{\frac{1}{2}{x}^3}-\underline{\underline{2x^2}}+\underline{\underline{\underline{3x}}}+\underline{\frac{1}{4}{x}^3}-\underline{\underline{\frac{3}{2}{x}^2}}-\underline{\underline{\underline{6x}}}+8=$   

$=\frac{3}{4}{x}^3-3\frac{1}{2}{x}^2-3x+8$