I'm pretty sure that this is correct?
$\displaystyle P(|z|=1) = P(z>1)+P(z<-1) = 2\times P(z>1) $
No, these are false...
What you did in a first time was right, except for the red part corrected by Mr F :
$\displaystyle \mathbb{P}(|Z|{\color{red}>}1)=\mathbb{P}(Z<-1)+\mathbb{P}(Z>1)$
Why so ? Because |Z|>1 means "Z<-1 OR Z>1"
Since the event "Z<-1" and the event "Z>1" cannot happen at the same time, $\displaystyle \mathbb{P}(|Z|>1)=\mathbb{P}(\{Z<-1\} \cup \{Z>1\})=\mathbb{P}(Z<-1)+\mathbb{P}(Z>1)$
(this is the formula of P(A u B)=P(A)+P(B) when A and B are disjoint)