Answer
(a) greater than
(b) $2.81diopters$
Work Step by Step
(a) We know that the near point distance is $67.0cm$ and $d_i=65.0cm$. As $d_{\circ}$ remains the same and the magnitude of $d_i$ is increased, therefore, the focal length of the lens decreases and the refractive power increases. Thus, the refractive power of the eye glasses is greater than $2.53$ diopters.
(b) We know that
$\frac{1}{f}=\frac{1}{d_{\circ}}+\frac{1}{d_i}$
We plug in the known values to obtain:
$\frac{1}{f}=\frac{1}{23cm}+\frac{1}{-65.0cm}$
$\implies f=35.5998cm=0.355998m$
Now the refractive power is given as
$\frac{1}{f}=\frac{1}{0.355998m}=2.81diopters$
