Answer
a). $(2)$ diverging
b). $-1.1D$
Work Step by Step
Given for a near-sightedness person his far point is 90 cm.
a). The type of lens it will need to correct the problem is (2) diverging lens, as he is unable to see objects at infinity which a person with normal eye can see. Hence it is needed to bring the object from distance more than his far point to his far point so that his eye can focus it.
This can be achieved by diverging lens of proper power.
b). $do=infinity$
$di=-90cm$
Now, $P=\frac{1}{f}=\frac{1}{do}+\frac{1}{di}$
$\frac{1}{f}=\frac{1}{infinity}+\frac{1}{-90}$
or, $P=\frac{100}{-90}m=-1.1D$