1. If $(2 x-1)^{5}=a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+$ $a_{1} x+a_{0}$, then $a_{2}+a_{4}=$ $\qquad$
,$- 1 .-120$.
Let $x=0$, we get $a_{0}=-1$.
Let $x=1$, we get $a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=1$;
Let $x=-1$, we get
$-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0}=-243$.
Adding the last two equations gives $a_{4}+a_{2}+a_{0}=-121$.
Therefore, $a_{2}+a_{4}=-120$.