We have, $P(A \cap B)=P(A) . P(B), P(B)=\frac{2}{7}$.
and
$$
P\left(A \cup B^c\right)=0.8
$$
Consider, $P\left(A \cup B^c\right)=0.8$
$$
\begin{aligned}
& \Rightarrow \quad P\left(A^t \cap B\right)^c=0.8 \Rightarrow \mathrm{l}-P\left(A^t \cap B\right)=0.8 \\
& \Rightarrow \quad P\left(A^c \cap B\right)=0.2 \Rightarrow P\left(A^c\right) \cdot P(B)=0.2 \\
&
\end{aligned}
$$
$\left[\because A\right.$ and $B$ are independent events, therefore $A^c$ and $B$ are also independent]
$$
P\left(A^c\right)=0.2 \times \frac{7}{2}=0.7 \Rightarrow P(A)=0.3
$$
and So, $P(A \cap B)=P(A) \cdot(B)=\frac{3}{10} \cdot \frac{2}{7}=\frac{6}{70}$
Now, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$$
=\frac{3}{10}+\frac{2}{7}-\frac{6}{70}=\frac{21+20-6}{70}=\frac{35}{70}=\frac{1}{2}
$$