Answer
(a) greater than
(b) $65cm$
Work Step by Step
(a) We know that the refractive power is $2.75$ diopters and the object distance remains the same. As the refractive power is increased, the focal length decreases and a decrease in focal length causes the increase in near point. Thus, the person's near point is greater than $57.0cm$.
(b) We know that
$f=\frac{1}{refractive \space power}$
$\implies f=\frac{1}{2.75diopters}=36.36cm$
and $d_{\circ}=23cm$
Now $\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_{\circ}}$
$\implies \frac{1}{d_i}=\frac{1}{36.36cm}-\frac{1}{23cm}$
$\implies d_i=-63cm$
Thus, the far point of the person is $63cm+2cm=65cm$
