The ratio

${\displaystyle \frac{\mathrm{AB}}{\mathrm{OC}}=\frac{\mathrm{OB}-\mathrm{OA}}{\mathrm{OC}}=\frac{3}{4}-\frac{2}{5}=\frac{7}{20}.}$

(2) By similar triangles

${\displaystyle \frac{\mathrm{OB}}{\mathrm{BC}}=\frac{3}{1}\mathrm{so}\frac{\mathrm{OB}}{\mathrm{OC}}=\frac{3}{4}.}$

Again by similar triangles,

${\displaystyle \frac{\mathrm{OA}}{\mathrm{OX}}=\frac{4}{3}}$

where $X$ is the midpoint of $\mathrm{OC}$ so

${\displaystyle \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{4}{14}.}$

The ratio

${\displaystyle \frac{\mathrm{AB}}{\mathrm{OC}}=\frac{\mathrm{OB}-\mathrm{OA}}{\mathrm{OC}}=\frac{3}{4}-\frac{2}{7}=\frac{13}{28}.}$

(3) This is exactly the same as (1) By similar triangles

${\displaystyle \frac{\mathrm{OB}}{\mathrm{BC}}=\frac{3}{1}\mathrm{so}\frac{\mathrm{OB}}{\mathrm{OC}}=\frac{3}{4}.}$

Again by similar triangles,

${\displaystyle \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{2}{3}\mathrm{so}\frac{\mathrm{OA}}{\mathrm{AC}}=\frac{2}{5}.}$

The ratio

${\displaystyle \frac{\mathrm{AB}}{\mathrm{OC}}=\frac{\mathrm{OB}-\mathrm{OA}}{\mathrm{OC}}=\frac{3}{4}-\frac{2}{5}=\frac{7}{20}.}$

Alternatively:
(1) $\mathrm{AB}=\frac{7}{20}\sqrt{10}$, $\mathrm{OC}=\sqrt{10}$.

(2) $\mathrm{AB}=\frac{13\sqrt{1}3}{28}$, $\mathrm{OC}=\sqrt{1}3$.

(3) $\mathrm{AB}=\frac{7\sqrt{1}3}{20}$, $\mathrm{OC}=\sqrt{1}3$.