It is given that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{7}{12} \text { and } \mathrm{P}(\text { not } \mathrm{A} \text { or not } \mathrm{B})=\frac{1}{4}$.

$\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\right)=\frac{1}{4}$

$\Rightarrow P\left((A \cap B)^{\prime}\right)=\frac{1}{4} \quad\left[A^{\prime} \cup B^{\prime}=(A \cap B)^{\prime}\right]$

$\Rightarrow 1-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}$

$\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{3}{4}$               ……….. $(1)$

However, $\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})=\frac{1}{2} \cdot \frac{7}{12}=\frac{7}{24} $          ………. $(2)$

Here, $\frac{3}{4} \neq \frac{7}{24}$

$\therefore $ $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \neq \mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})$

Therefore, $A$ and $B$ are not independent events.