Going by options:
(a) Minimum value of $P(A\cap B)$ is the higher of $P(A)$ and $P(B)$ i.e. \dfrac { 2 }{ 3 }  and maximum value is $1$. So, (a) is true.
(b) Maximum intersection of $P(A\cap \bar{B})$ is the minimum of $P(A)$ and P(\bar { B } ) i.e. minimum of  \dfrac { 1 }{ 2 }  and \dfrac { 1 }{ 3 }  = \dfrac { 1 }{ 3 } . So, (b) is true.
(c) $P(A)+P(B)={\displaystyle \frac{1}{2}}+{\displaystyle \frac{2}{3}}={\displaystyle \frac{7}{6}}$. Hence, P(A\cap B)\ge \dfrac { 7 }{ 6 } -1=\dfrac { 1 }{ 6 } . 
Again, maximum intersection of $P(A\cap B)$ is the minimum of $P(A)$ and $P(B)$ i.e. minimum of  \dfrac { 1 }{ 2 }  and \dfrac { 2 }{ 3 }  = \dfrac { 1 }{ 2 } . So, (c) is true.
(d) Maximum intersection of $P(B\cap \bar{A})$ is the minimum of $P(B)$ and P(\bar { A } ) i.e. minimum of  \dfrac { 2 }{ 3 }  and \dfrac { 1 }{ 2 }  = \dfrac { 1 }{ 2 } .
Again, $P(\bar{A})+P(B)={\displaystyle \frac{1}{2}}+{\displaystyle \frac{2}{3}}={\displaystyle \frac{7}{6}}$. Hence, P(\bar {A}\cap B)\ge \dfrac { 7 }{ 6 } -1=\dfrac { 1 }{ 6 } . So, (d) is true.
Hence, (a), (b), (c), (d) are correct.