[tex]x=-2,\ \ \ y=-1\\\\a)\\\\\frac{1}{3}x^3y=\frac{1}{3}*(-2)^3*(-1)=\frac{1}{3}*(-8)*(-1)=\frac{8}{3}=2\frac{2}{3}\\\\b)\\\\-\frac{1}{2}xy^2=-\frac{1}{\not{2}^1}*(- \not{2}^1 )*(-1)^2=1*1=1\\\\c)\\\\2 \frac{3}{4}(x^2-y^2)*x^2y= \frac{11}{4}*( (-2)^2- (-1)^2)* (-2)^2*(-1)=\\\\=\frac{11}{\not{4}^1}*(4-1)*\not{4}^1*(-1)=11*3*(-1)=-33[/tex]
