I have solved an exercise but i have no idea if i am right or wrong.
Find $P(A\cup B')$ and $P(A'\cup B)$ when $A,B$ are independent and $P(A)=0.2,\,P(B)=0.8.$
First I use $$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$ So now $\;0.2+0.8-0.06=0.94.$ After this, i think there is a type (i wasn't sure for this):
$$P(A)=1-P(A') \;\implies\; P(A')= -P(A)+1\;=-0.2+1\;=0.8,$$ $$P(B)=1-P(B') \;\implies\; P(B')= -P(B)+1\;=-0.8+1\;=0.2.$$
As a result to find what it asks me, i did this, but i am not sure if i am right on everything i did:
$$P(A\cup B')=P(A)+P(B')-P(A\cap B')= 0.2+0.2-(0.2)^2=\dots.$$ I did the same for $$P(A'\cup B)=P(A')+P(B)-P(A'\cap B)=\dots.$$
Note that
$$P(A \cup B)=P(A)+P(B)-P(AB)=P(A)+P(B)-P(A)P(B)=0.2+0.8-0.16=0.84$$
Your working seems fine but be careful of careless mistakes.
Note that independence is used so that we can compute the intersection probability $P(AB)=P(A)P(B)$.