Yes, since a + z = a for all a in F, we have a = 0 in particular that yields 0 + z = 0.

Similar situation occuring in your second question. The axiom is a + 0 = a for all a in F. So in particualr when a = z, we see that z + 0 = z. a is any arbitrary element and z is a specific unknown element (which we soon shall see is zero).

I hope this helps.