Finding how many 8-bit bytes contain an even number of zeros . . .

Any $7$-bit word can be completed to an $8$-bit word with an even number of $0$'s in exactly one way by choosing the eighth bit suitably. So the number of $8$-bit words with an even number of $0$'s is the same as the number of $7$-bit words. This is $2^7$.

HINT

Divide the cases for the number of zeros, namely $0,2,4,6,8$, which gives us $$\binom{8}{0}+\binom{8}{2}+\binom{8}{4}+\binom{8}{6}+\binom{8}{8}=\frac{2^8}{2}$$ From the Binomial Theorem.

You could also use recurrence relations to solve this problem.

There are $2^8 = 256$ different bit-strings of ones and zeros. Because of symmetry$^{(*)}$ the number of strings with an odd number of zeros must be the same as the number of strings with an even number of zeros. Specifically, half of all the possible cases have an even number of zeros, so we have have $\frac{1}{2}2^8 = 2^7$ such bit-strings.

$(*)$ The symmetry I am thinking of here is the symmetry between an even and odd count of zeros. Because we are considering all bit-strings of length eight, if we are asking how many of them have an even/odd number of zeros, there is no reason to favor even or odd over the other, so there must be the same number of each.