(a) It is the case of hypermetropia.
For a hypermetropic eye, $\mathit{u}=-25\mathrm{cm}$ and $\mathit{v}=-50\mathrm{cm}$
Using lens formula
$\frac{1}{\mathit{f}}=\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}$
$\frac{1}{\mathit{f}}=-\frac{1}{50}+\frac{1}{25}$
$\frac{1}{\mathit{f}}=\frac{1}{50}$
We get focal length is $\mathit{f}=50\mathrm{cm}=0.5\mathrm{m}$
From the formula, $\mathrm{P}=\frac{1}{\mathit{f}}$
$\mathrm{P}=\frac{1}{0.5}=2\mathrm{D}$
So the power of the lens is 2 dioptre and the nature of lens is a converging lens or a convex lens.