$\text{a)}\left(3x^2+2x-4\right)x^2=3x^2\cdot x^2+2x\cdot x^2-4\cdot x^2=3x^4+2x^3-4x^2$      

$\text{b)}-2x^3\left(\frac{1}{4}{x}^2-\frac{1}{2}x-2\right)=-2x^3\cdot \frac{1}{4}{x}^2-2x^3\cdot \left(-\frac{1}{2}x\right)-2x^3\cdot \left(-2\right)=$  

$=-\frac{1}{2}{x}^5+x^4+4x^3$  

$\text{c)}-\frac{1}{2}{x}^2\left(4x^{2}y-2x{y}^2\right)=-\frac{1}{2}{x}^2\cdot 4x^{2}y-\frac{1}{2}{x}^2\cdot \left(-2x{y}^2\right)=-2x^{4}y+x^3{y}^2$  

$\text{d)}\sqrt{2}{x}^4\left(\sqrt{8}xy+y^2\right)=\sqrt{2}{x}^4\cdot \sqrt{8}xy+\sqrt{2}{x}^4\cdot y^2=$  

$=\sqrt{16}{x}^{5}y+\sqrt{2}{x}^4{y}^2=4x^{5}y+\sqrt{2}{x}^4{y}^2$  

$\text{e)}\left(x^{2}y-xy+x{y}^2\right)x^{2}y=x^{2}y\cdot x^{2}y-xy\cdot x^{2}y+x{y}^2\cdot x^{2}y=x^4{y}^2-x^3{y}^2+x^3{y}^3$  

$\text{f)}2\sqrt{3}x{y}^3\left(\sqrt{3}{x}^2{y}^2-\frac{1}{2}{x}^{3}y\right)=2\sqrt{3}x{y}^3\cdot \sqrt{3}{x}^2{y}^2+2\sqrt{3}x{y}^3\cdot \left(-\frac{1}{2}{x}^{3}y\right)=$  

$=2\sqrt{9}{x}^3{y}^5-\sqrt{3}{x}^4{y}^4=6x^3{y}^5-\sqrt{3}{x}^4{y}^4$