The ratio of the greatest value of $2-\cos \mathrm{x}+{\sin }^2\mathrm{x}$ is to its least value, is 

Let,$\mathrm{y}=2-\cos \mathrm{x}+{\sin }^2\mathrm{x}$ then,

$\begin{array}{rl}\mathrm{y}& =2-\cos \mathrm{x}+1-{\cos }^2\mathrm{x}\\ \Rightarrow \mathrm{y}& =3-({\cos }^2\mathrm{x}+\cos \mathrm{x})\Rightarrow \mathrm{y}=\frac{13}{4}-{(\cos \mathrm{x}+\frac{1}{2})}^2\end{array}$

Now,

$\begin{array}{rl} & -1\le \cos \mathrm{x}\le 1\text{for all}\mathrm{x}\\ \Rightarrow & -\frac{1}{2}\le (\cos \mathrm{x}+\frac{1}{2})\le \frac{3}{2}\text{for all}\mathrm{x}\\ \Rightarrow & 0\le {(\cos \mathrm{x}+\frac{1}{2})}^2\le \frac{9}{4}\text{for all}\mathrm{x}\\ \Rightarrow & -\frac{9}{4}\le -{(\cos \mathrm{x}+\frac{1}{2})}^2\le 0\text{for all}\mathrm{x}\\ \Rightarrow & \frac{13}{4}-\frac{9}{4}\le \frac{13}{4}-{(\cos \mathrm{x}+\frac{1}{2})}^2\le \frac{13}{4}\text{for all}\mathrm{x}\\ \Rightarrow & 1\le \mathrm{y}\le \frac{13}{4}\\ \therefore & {\mathrm{y}}_{max}=\frac{13}{4}\text{and}{\mathrm{y}}_{min}=1\end{array}$

Hence, required ratio is $\frac{13}{4}$