Probability that A solves the problem $=\frac{1}{2}$
$\Rightarrow$ Probablility that A does not solve the problem $P(\bar{A})=1-\frac{1}{2}=\frac{1}{2}$
Probability that $\mathrm{B}$ solves the problem $=\frac{1}{3}$
$\Rightarrow$ Probability that $\mathrm{B}$ does not solved the problem $=1-\frac{1}{3}=\frac{2}{3}$
(i) Probability that problem is not solved $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}})=\frac{1}{2} \times \frac{2}{3}=\frac{1}{3}$ $\Rightarrow$ Probability that problem is solved $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=1-\frac{1}{3}=\frac{2}{3}$
(ii) Exactly one of then solves the problem $=P(A \cap \bar{B})+P(\bar{A} \cap B)$ $=\mathrm{P}(\mathrm{A}) \mathrm{P}(\overline{\mathrm{B}})+\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\mathrm{B})$
Since $\mathrm{A}$ and $\mathrm{B}$ are independent events $\mathrm{So} \overline{\mathrm{A}} \cap \mathrm{B}$ and $\mathrm{A} \cap \overline{\mathrm{B}}$ are also independent
Now
$$
\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{2}{3}, \mathrm{P}(\overline{\mathrm{A}})=\frac{1}{2}, \mathrm{P}(\overline{\mathrm{B}})=\frac{1}{3}
$$
Exactly one of them solves the problem
$$
=\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}=\frac{1}{3}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}
$$