Hint: The image formed by the lens acts as an object for the eye-lens.

Step 1: Find the power of the glasses.

(i) For the myopic person,

Object distance, $\mathrm{u}=-0.1 \mathrm{m}$

Image distance, $\mathrm{v}=+0.02 \mathrm{m}$

The power, ${\mathrm{P}}_{\mathrm{f}}=\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{0.02}+\frac{1}{0.1}=60 \mathrm{D}$
By the corrective lens, the object distance at the far point is $\infty$.
The required power is,
                            $\mathrm{P}{\text{'}}_{\mathrm{f}}=\frac{1}{\mathrm{f}\text{'}}=\frac{1}{\infty }+\frac{1}{0.02}=50 \mathrm{D}$
The power of the glasses,
${\mathrm{P}}_{\mathrm{g}}=50 \mathrm{D}-60\hspace{0.17em}\mathrm{D}=-10 \mathrm{D}$

Step 2: Find the near-point without glasses.

(ii) His power of accommodation is 4 D for the normal eye. Let the power of the normal eye for near vision be ${\mathrm{P}}_{\mathrm{n}}$.
Then,                          $4={\mathrm{P}}_{\mathrm{n}}-{\mathrm{P}}_{\mathrm{f}} \mathrm{or} {\mathrm{P}}_{\mathrm{n}}=64 \mathrm{D}$
Let his near point be ${\mathrm{x}}_{\mathrm{n}}$, then,
                                           $\frac{1}{{\mathrm{x}}_{\mathrm{n}}}+\frac{1}{0.02}=64 \mathrm{or} \frac{1}{{\mathrm{x}}_{\mathrm{n}}}+50=64$
                                           ${\mathrm{x}}_{\mathrm{n}}=\frac{1}{14}$  
$\therefore$                                        ${\mathrm{x}}_{\mathrm{n}}=\frac{1}{14}=0.07 \mathrm{m}$
Step 2: Find the near-point with glasses.
(iii) With glasses $\mathrm{P}{\text{'}}_{\mathrm{n}}=\mathrm{P}{\text{'}}_{\mathrm{f}}+4=54$
                                             $54=\frac{1}{\mathrm{x}{\text{'}}_{\mathrm{n}}}+\frac{1}{0.02}=\frac{1}{\mathrm{x}{\text{'}}_{\mathrm{n}}}+50$
$\frac{1}{\mathrm{x}{\text{'}}_{\mathrm{n}}}=4$

$\therefore$                                        $\mathrm{x}{\text{'}}_{\mathrm{n}}=\frac{1}{4}= 0.25 \mathrm{m}$