Answer
a). $-340X$
b). $3900$%
Work Step by Step
a). $fo=0.5cm$
$fe=3.25m$
$L=22cm$
Also, $m_{total}=-\frac{25(cm)L}{fofe}=-\frac{25\times22}{0.5\times3.25}=-340X$
b). To compare total magnification with that of eyepiece alone as magnifying glass in percentage :-
$m_{e}=1+\frac{25}{fe}=1+\frac{25}{3.25}=8.69X$
Hence, the required percentage = $\frac{m_{total}-m_{e}}{m_{e}}\times 100=\frac{340-8.69}{8.69}\times 100=3900$%