$P(A \cup B) = P(A \cap B)$ if and only if the relation between $P(A)$ and $P(B)$ is

Two dice are thrown independently. Let $A$ be the event that the number appeared on the $1^{\text {st }}$ die is less than the number appeared on the $2^{\text {nd }}$ die, $B$ be the event that the number appeared on the $1^{\text {st }}$ die is even and that on the second die is odd, and $C$ be the event that the number appeared on the $1^{\text {st }}$ die is odd and that on the $2^{\text {nd }}$ is even. Then

Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(A \cup B)$

Let $X$ and $Y$ are two events such that $P(X \cup Y=P)\,(X \cap Y).$

Statement $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$

Statement $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$

Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $P(B)=p .$ Find $p$ if they are independent.

A coin is tossed twice. If events $A$ and $B$ are defined as :$A =$ head on first toss, $B = $ head on second toss. Then the probability of $A \cup B = $