1. Homogeneización de la primera ecuación:
Usamos $\sin^2 x = \frac{1-\cos 2x}{2}$ y $\cos^2 y = \frac{1+\cos 2y}{2}$:
$$ \frac{1-\cos 2x}{2} + \frac{1+\cos 2y}{2} = \frac{1}{2} \implies 2 - \cos 2x + \cos 2y = 1 $$
$$ \cos 2y - \cos 2x = -1 $$

2. Transformación de resta de cosenos:
$$ 2 \sin(x+y) \sin(x-y) = -1 $$
Sustituyendo $x+y = \frac{\pi}{4}$:
$$ 2 \sin\left(\frac{\pi}{4}\right) \sin(x-y) = -1 \implies 2 \left(\frac{\sqrt{2}}{2}\right) \sin(x-y) = -1 $$
$$ \sin(x-y) = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} $$

3. Soluciones:
$x-y = -\frac{\pi}{4} + 2k\pi$ o $x-y = \frac{5\pi}{4} + 2k\pi$.
Combinando con $x+y = \frac{\pi}{4}$:
Para $x-y = -\frac{\pi}{4} \implies 2x = 0 \implies x = 0, y = \frac{\pi}{4}$.

Resultado final:
$$ \boxed{x = k\pi, \quad y = \frac{\pi}{4} - k\pi} \text{ (considerando periodicidad)} $$