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$\frac{(2+3i)(5-i)}{2+i}= \frac{10+15i-2i-3i^2}{2+i}= \frac{10+13i+3}{2+i}= \frac{13+13i}{2+i}= \frac{(13+13i)(2-i)}{(2+i)(2-i)}=\\\\= \frac{26+26i-13i-13i^2}{4-i^2}= \frac{26+13i+13}{4+1}= \frac{39+13i}{5}=7,8+2,6i$

$(2i-1)^4-((2i+1)^4=((2i-1)^2+(2i+1)^2)((2i-1)^2-(2i+1)^2)=\\\\=(4i^2-4i+1+4i^2+4i+1)(4i^2-4i+1-4i^2-4i-1)=\\\\=(-4-4+2)(-8i)=-6*(-8i)=48i$