$\frac{sin^4x+cos^4x+0,5sin^22x}{2cos2x+2sin^2x}$$=\frac{sin^4(\frac\pi3)+cos^4(\frac\pi3)+0,5sin^22(\frac\pi3)}{2cos2(\frac\pi3)+2sin^2(\frac\pi3)}=$

$=\frac{\frac9{16}+\frac1{16}+0,5\cdot\frac34}{2\cdot\left(-0,5\right)+2\cdot\frac34}=\frac1{\frac12}=2$