$a)2\left(x-5\right)\cdot 3\left(x^2+4\right)=2\cdot 3\left(x^3+4x-5x^2-20\right)=6\left(x^3-5x^2+4x-20\right)=6x^3-30x^2+24x-120$  

b) W liczniku będizemy mieli wzór na różnicę sześcianów.

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

$c)\left(x-3\right)\left(x^2+2\right)\left(3+x\right)\left(2-x^2\right)=\left(x-3\right)\left(x+3\right)\left(2+x^2\right)\left(2-x^2\right)=\left(x^2-9\right)\left(4-x^4\right)=4x^2-x^6-36+9x^4=-x^6+9x^4+4x^2-36$  

$d)4{\left(x+4\right)}^2{\left(\frac{1}{2}x-2\right)}^2={\left(2\left(x+4\right)\left(\frac{1}{2}x-2\right)\right)}^2={\left(\left(x+4\right)\left(x-4\right)\right)}^2={\left(x^2-16\right)}^2=x^4-32x^2+256$