I can understand the second statement from the Pauli matrices... However I think that I don't understand the 1st statement as it is... why would the diagonal elements of an odd-operator be zero if parity is definite?

If ##A## is an odd operator and ##\pi## is a parity operator, then we have ##\pi^\dagger A \pi = -A##. If furthermore a state ##|\psi\rangle## which has a definite parity sandwhiches the expression ##\pi^\dagger A \pi = -A## from left and right in each side of the equation, you will get ##\langle \psi| A |\psi \rangle = -\langle \psi|A|\psi \rangle## because ##\langle \psi|\pi^\dagger A \pi|\psi \rangle = \langle \psi| A |\psi \rangle##. You see then we must require ##\langle \psi| A |\psi \rangle = 0## so that the previous relation can hold.