To prove (1):
We know that
$$|\cos(\alpha + \beta)| = |\cos \alpha \cos \beta - \sin \alpha \sin \beta| \leq |\cos \alpha \cos \beta| + |\sin \alpha \sin \beta|$$
Using the property that $|ab| \leq |a| |b|$ for any real numbers $a$ and $b$, we have
$$|\cos \alpha \cos \beta| \leq |\cos \alpha| \quad \text{and} \quad |\sin \alpha \sin \beta| \leq |\sin \beta|$$
Therefore, $|\cos(\alpha + \beta)| \leq |\cos \alpha| + |\sin \beta|$, as desired.

Similarly,
$$|\sin(\alpha + \beta)| = |\sin \alpha \cos \beta + \cos \alpha \sin \beta| \leq |\sin \alpha \cos \beta| + |\cos \alpha \sin \beta|$$
and since $|a| + |b| \geq |a+b|$, we get
$$|\sin \alpha \cos \beta| + |\cos \alpha \sin \beta| \leq |\cos \alpha| + |\cos \beta|$$
Thus, $|\sin(\alpha + \beta)| \leq |\cos \alpha| + |\cos \beta|$, which completes the proof for part (1).

To prove (2):
From part (1), we can apply the result to $\alpha + (\beta + \gamma)$ since $\alpha + \beta + \gamma = 0$. This gives us
$$|\cos(\alpha + \beta + \gamma)| \leq |\cos \alpha| + |\sin (\beta + \gamma)|$$
$$|\cos(\alpha + \beta + \gamma)| = |\cos 0| = 1$$
On the other hand,
$$|\sin (\beta + \gamma)| = |\sin (-\alpha)| = |\sin \alpha| \leq |\cos \beta| + |\cos \gamma|$$
by applying part (1) to $\beta + \gamma = -\alpha$.

Combining these inequalities gives us:
$$|\cos \alpha| + |\cos \beta| + |\cos \gamma| \geq |\cos \alpha| + |\sin (\beta + \gamma)| = |\cos(\alpha + \beta + \gamma)| = 1$$
Hence,
$$|\cos \alpha| + |\cos \beta| + |\cos \gamma| \geq \boxed{1}$$
which is what we wanted to prove.