Frequency response simple question

You may be confused about how exponentials interact.

The important one for this is \$e^x \cdot e^x = e^{x+x} = e^{2x}\$

Take your equations and work them backwards to see what happened. You have

\$H(\omega)=0.5e^\frac{-j\omega}{2}(e^\frac{j\omega}{2}+e^\frac{-j\omega}{2})\$

Distribute the \$e^\frac{-j\omega}{2}\$ term inside the parentheses and combine the terms

\$(e^\frac{-j\omega}{2} \cdot e^\frac{j\omega}{2} + e^\frac{-j\omega}{2} \cdot e^\frac{-j\omega}{2}) = (e^{\frac{-j\omega}{2}+\frac{j\omega}{2}} + e^{\frac{-j\omega}{2}+\frac{-j\omega}{2}}) = (e^0 + e^\frac{-2j\omega}{2})=(1+e^{-j\omega})\$

So adding the rest of the equation you have

\$ H(\omega) = 0.5(1+e^{-j\omega})\$

In other words, that \$e\$ term was divided-out/factored-out of the parentheses.

They are factoring out \$e^\frac{-j\omega}{2}\$.

Note that \$e^\frac{-j\omega}{2} \cdot e^\frac{j\omega}{2} = 1\$

and \$e^\frac{-j\omega}{2} \cdot e^\frac{-j\omega}{2} = e^{-j\omega}\$