(A) The defect of vision is Hypermetropia (farsightedness).
$(b)$ Two reasons for this defect are:
$(i)$ The eyeball has become too short,so that light rays from a nearby object are focused at a point behind the retina.
$(ii)$ The focal length of the eye lens is too long because the ciliary muscles are unable to make the lens sufficiently convex.
$(c)$ Given: Object distance $u = -25\, cm$,Image distance $v = -50\, cm$ (the near point of the defective eye).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25} = \frac{-1 + 2}{50} = \frac{1}{50}$
$f = +50\, cm = +0.5\, m$
Power $P = \frac{1}{f(m)} = \frac{1}{0.5} = +2.0\, D$.
The nature of the lens is a convex lens.
$(d)$ The ray diagrams are as shown in the image.