Maximum: We note that $A\cap B\subseteq A$. Frommonotonicity, it follows that

$$P(A\cap B)\leq P(A)=0.4 $$

To show that 0.4 is the maximum, we still have to show that it is possible to attain 0.4, which is normally done by contructing an example. It may be easiest to draw a picture, but we can do it using symbols as follows:

Let sample space $\Omega = \{1,2,3,4,5,6,7,8,9,10\}$ with probability function that gives probability $1/10$ to every outcome. Let $A = \{1,2,3,4\}$ and $B = \{1,2,3,4,5,6,7\}$. Then $P(A) = 0.4$ and $P(B) = 0.7$ as required, and

$$P(A \cap B) = P( \{1,2,3,4\} ) = P( \{1\} ) + P( \{2\} ) + P( \{3\} ) + P( \{4 \} ) = 1/10 + 1/10 + 1/10 + 1/10 = 0.4$$

Since 0.4 is possible, but no probability above 0.4 is possible, 0.4 is the maximum.

Minimum: From the additive rule, we have

$$P(A\cap B)=P(A)+P(B)-P(A\cup B)$$

Since the last probability is at most 1, we can write

$$P(A\cap B) \ge P(A)+P(B)-1 =0.4+0.7-1=0.1$$

To show that it is possible to attain 0.1, let sample space $\Omega = \{1,2,3,4,5,6,7,8,9,10\}$ with probability function that gives probability $1/10$ to every outcome. Let $A = \{1,2,3,4\}$ and $B = \{4,5,6,7,8,9,10\}$. Then $P(A) = 0.4$ and $P(B) = 0.7$ as required, and

$$P(A \cap B) = P( \{4\} ) = 0.1$$

Since 0.1 is possible, but no probability below 0.1 is possible, 0.1 is the minimum.