Given $P(A)$ and $P(B)$, calculate the maximum and minimum values of $P(A \cup B)$ and $P(A \cap B)$ I considered two cases: $A$ and $B$ are disjoint and $A$ and $B$ are not disjoint.
In case $A$ and $B$ are disjoint, the maximum value $P(A \cup B)$ attains is $P(A) + P(B)$ and the minimum value $P(A \cap B)$ attains is $0$.
In case $A$ and $B$ are not disjoint, the value $P(A \cup B)$ attains is greater than the greater of $P(A)$ and $P(B)$ and the value $P(A \cap B)$ attains is less than the lesser of $P(A)$ and $P(B)$.
I'm not sure if this is the right logic, but I can't think of any other cases.
 A: It is evident that$$P(A\cap B)\leq\min(P(A),P(B))\tag1$$and if one of the sets $A,B$ is a subset of the other then we have equality.
Also it is evident that $$P(A\cup B)\leq\min(1,P(A)+P(B))\tag2$$
and also here equality can arise if $A,B$ are disjoint, or - if that is not possible - if $A\cup B=\Omega$.
Making use of: $$P(A\cap B)+P(A\cup B)=P(A)+P(B)$$ we find the minima:$$P(A\cap B)\geq P(A)+P(B)-\min(1,P(A)+P(B))=\max(P(A)+P(B)-1,0)$$ and:$$P(A\cup B)\geq P(A)+P(B)-\min(P(A),P(B))=\max(P(A),P(B))$$
A: The basic formula is $$P(A \cup B)=P(A)+P(B)-P(A\cap B).$$
So, if $A$ and $B$ are disjoint:


*

*$P(A\cap B)$ is always $0$ (by definition of disjointness);

*So, according to the formula, $P(A\cup B)$ must be $P(A)+P(B)$.


If $A$ and $B$ are not disjoint, $P(A\cap B)\geqslant0$ and:


*

*$P(A\cap B)\leqslant \min (P(A), P(B))$ (since both $A\cap B\subseteq A$, so $P(A\cap B)\leqslant P(A)$, and $A\cap B\subseteq B$, so $P(A\cap B)\leqslant P(B)$)

*It follows that $P(A\cup B)\geqslant P(A)+P(B)-\min(P(A), P(B))=\max(P(A), P(B))$
(obviously $\forall x, y : \min(x, y)+\max(x, y)=x+y$)

A: For Max $P(A\cap B)$:
$$\begin{cases}P(A\cap B)=P(A)\cdot P(B|A)\le P(A)\\ 
P(A\cap B)=P(B)\cdot P(A|B)\le P(B)\end{cases} \Rightarrow \\
\color{red}{\text{Max }P(A\cap B)}= \text{Min}\{P(A),P(B)\}.$$
Note: 
$$1) \ \text{the equality occurs when $P(B|A)=1$ (i.e. $B\subseteq A$) or $P(A|B)=1$ (i.e. $A\subseteq B$).} \\
2) \ \begin{cases}x\le 0.5 \\ x\le 0.3\end{cases} \Rightarrow x\le 0.3.$$
For Max $P(A\cup B)$:
$$\begin{cases}P(A\cup B)=P(A)+P(B)-P(A\cap B)\le P(A)+P(B)\\
P(A\cup B)\le 1\end{cases} \Rightarrow \\
\color{blue}{\text{Max }P(A\cup B)}=\text{Min} \{1,P(A)+P(B)\}.$$
Note: 
$$1) \ \text{the equality occurs when $P(A\cap B)=0$ (i.e. $A\cap B=\emptyset$) or $A\cup B=S$}. \\
2) \ \begin{cases}x\le 0.5 \\ x\le 0.3\end{cases} \Rightarrow x\le 0.3.$$
For Min $P(A\cap B)$:
$$\begin{align}\text{Min }P(A\cap B)&=P(A)+P(B)-\color{blue}{\text{Max }P(A\cup B)}\Rightarrow \\
&=\text{Max }\{P(A)+P(B)-1,0\}.\end{align}$$
Note:
$$\begin{cases}P(A\cap B)\ge P(A)+P(B)-1\\ P(A\cap B)\ge P(A)+P(B)-(P(A)+P(B))=0\end{cases} \ \text{and}\\
\begin{cases}x\ge 0.5 \\ x\ge 0.3\end{cases} \Rightarrow x\ge 0.5.$$
For Min $P(A\cap B)$:
$$\begin{align}\text{Min }P(A\cup B)&=P(A)+P(B)-\color{red}{\text{Max }P(A\cap B)}=\\
&=\text{Max }\{P(A),P(B)\}.\end{align}$$
Note:
$$\begin{cases}P(A\cup B)\ge P(A)+P(B)-P(A)=P(B)\\ P(A\cup B)\ge P(A)+P(B)-P(B)=P(A)\end{cases} \ \text{and}\\
\begin{cases}x\ge 0.5 \\ x\ge 0.3\end{cases} \Rightarrow x\ge 0.5.$$
