Five in Three

Let's assume the pentagon has side length 1.

Solve blue $\triangle BGC$ . Its angles are easily deduced. $\hspace{.2cm}\angle{G} = {60}^\circ$ and $\angle{C} = {48}^\circ$

By the law of sines, $\overline{BG} = \frac{\sin{48}^\circ}{\sin{60}^\circ} \approx .858$

Since $\triangle{FAE}$ and $\triangle{BGC}$ are congruent, $\hspace{.2cm}\overline{FA} = \overline{BG}$ and $\hspace{.2cm}\overline{FG} = 2\times\overline{BG} + 1 \approx 2.716$

So $\triangle{FGH} = \frac{\sqrt{3}}{4}\times\overline{FG}^2 \approx 3.195$

The red area of the pentagon is $\frac{5}{4}\times$ $\sqrt{ 1 + \frac{2}{\sqrt{5} }}$ $\times \overline{AB}^2 \approx 1.72$

Thus the total blue area is $3.195 - 1.72 \approx 1.475 < 1.72$

Red wins.