Which redexes are there in $\lambda s. \lambda z. (\lambda u. z)(\lambda v. v)$? How to substitute arguments?
You must learn how to put in parentheses and then it will be easier to figure out what is what. In the above case, we first put in parentheses:
$$\lambda s . (\lambda z . ((\lambda u. z) (\lambda v . v))).$$
This is the only correct way to put back parentheses. For instance, this is wrong
$$(\lambda s . (\lambda z . (\lambda u . z))) (\lambda v . v)$$
Why is it wrong? Because the rules for writing expressions without parentheses say that when you see $\lambda x . \cdots$ that means that $\lambda x$ binds the whole expression. For instance $\lambda x . x (\lambda y . y)$ is the same as $\lambda x . (x (\lambda y . y))$ and is diffrerent from $(\lambda x . x) (\lambda y . y)$.

If this is not resolving your dilemma you have to explain why you think that the other redex is valid.