We are given that ${\displaystyle P(A\cup B)=0.6,}$ and ${\displaystyle P(A\cap B)=\mathrm{0.2.}}$
We know that if A and B are any two events, then
${\displaystyle P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)}$
or ${\displaystyle 0.6=1-P\left(\bar{A}\right)+1-P\left(\bar{B}\right)-0.2}$
or ${\displaystyle P\left(\bar{A}\right)+P\left(\bar{B}\right)=2-0.2-0.6=\mathrm{1.2.}}$