Let $\left(\mathit{x}-2\right)$ is a factor of the given expression.
$\mathit{x}-2=0$
$\mathit{x}=2$
Given: $2{\mathit{x}}^3+\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}-14=0$
$2{\left(2\right)}^3+\mathit{a}{\left(2\right)}^2+\mathit{b}\left(2\right)-14=0$
$16+4\mathit{a}+2\mathit{b}-14=0$
$4\mathit{a}+2\mathit{b}+2=0$
$4\mathit{a}+2\mathit{b}=-2$

and when given expression is divided by $\left(\mathit{x}-3\right)$ .
$\mathit{x}-3=0$
$\mathit{x}=3$
$\therefore 2{\mathit{x}}^3+\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}-14=52$
$2{\left(3\right)}^3+\mathit{a}{\left(3\right)}^2+\mathit{b}\left(3\right)-66=0$
$54+9\mathit{a}+3\mathit{b}-66=0$
$9\mathit{a}+3\mathit{b}=12$
$3\mathit{a}+\mathit{b}=4$ ............(ii)
Solving equation (i) and (ii),
$2\mathit{a}+\mathit{b}=-1$
$3\mathit{a}+\mathit{b}=4$
$(-)(-)(-)$
$\mathit{─}\mathit{─}\mathit{─}\mathit{─}\mathit{─}\mathit{─}\mathit{─}\mathit{─}$
$-\mathit{a}=-5\Rightarrow \mathit{a}=5$
from (ii),
$3\times 5+\mathit{b}=4$
$\mathit{b}=4-15=-11$