1. Datos y sistema:
1) $\sin \alpha + \sin \beta = a$
2) $\cos \alpha + \cos \beta = b$

Transformando a producto:
$$ 2 \sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right) = a $$
$$ 2 \cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right) = b $$

2. Desarrollo:
Dividiendo (1) entre (2):
$$ \tan\left(\frac{\alpha+\beta}{2}\right) = \frac{a}{b} $$

Sea $\theta = \frac{\alpha+\beta}{2}$, entonces $\tan \theta = \frac{a}{b}$. Buscamos $\cos 2\theta$ y $\sin 2\theta$:
Para (i) $\cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$:
$$ \cos(\alpha + \beta) = \frac{1 - (a/b)^2}{1 + (a/b)^2} = \frac{\frac{b^2 - a^2}{b^2}}{\frac{b^2 + a^2}{b^2}} = \frac{b^2 - a^2}{b^2 + a^2} $$

Para (ii) $\sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}$:
$$ \sin(\alpha + \beta) = \frac{2(a/b)}{1 + (a/b)^2} = \frac{\frac{2a}{b}}{\frac{b^2 + a^2}{b^2}} = \frac{2ab}{b^2 + a^2} $$

3. Conclusión:
Las identidades quedan demostradas.
$$ \boxed{\cos(\alpha + \beta) = \frac{b^2 - a^2}{b^2 + a^2}, \quad \sin(\alpha + \beta) = \frac{2ab}{b^2 + a^2}} $$