(a)
Hint: Use the lens formula.

Step 1: Find the closest distance at which the child can read the book.
Given that:

The focal length of the magnifying glass, f=5 cm

Let the shortest object distance = u

When the child sees through the lens placed very close to its eye, the image of the fonts should be made at the near point of the child. So,

Image distance, v=-d=-25 cm

According to the lens formula,

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$
$\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{f}}$
$=\frac{1}{-25}-\frac{1}{5}=\frac{-5-1}{25}=\frac{-6}{25}$
$\therefore \mathrm{u}=-\frac{25}{6}=-4.167 \mathrm{cm}$

Hence, the closest distance at which the child can read the book is 4.167 cm.

Step 2: Find the longest distance at which the child can read the book.
When the object is at the longest distance (u’), the image distance should be made at infinity.

According to the lens formula,

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}\text{'}}-\frac{1}{\mathrm{u}\text{'}}$
$\frac{1}{\mathrm{u}\text{'}}=\frac{1}{\infty }-\frac{1}{5}=-\frac{1}{5}$
$\therefore \mathrm{u}\text{'} =- 5 \mathrm{cm}$

Hence, the farthest distance at which the child can read the book is 5 cm.

(b)
Hint: \(\alpha =\frac{d}{u}\)
Step 1: Find the maximum angular magnification.
Maximum angular magnification:

${\mathrm{\alpha }}_{\max }=\frac{\mathrm{d}}{\left|\mathrm{u}\right|}=\frac{25}{{\displaystyle \frac{25}{6}}}=6$

Step 2: Find the minimum angular magnification.
Minimum angular magnification:

${\mathrm{\alpha }}_{\max }=\frac{\mathrm{d}}{\left|\mathrm{u}\text{'}\right|}=\frac{25}{5}=5$