Answer
(a) farther away
(b) $34cm$
Work Step by Step
(a) As $d_i=\frac{1}{\frac{1}{f}-\frac{1}{d_{\circ}}}$
We know that $m=-\frac{d_i}{d_{\circ}}$
$\implies m=-\frac{1}{d_{\circ}(\frac{1}{f}-\frac{1}{d_{\circ}})}$
$\implies m=\frac{1}{1+\frac{d_{\circ}}{f}}$
Thus, if we want to decrease the magnification, the value of $1+\frac{d_{\circ}}{f}$ should be increased -- that is, the distance of the object should increase.
(b) We know that
$d_{\circ \space initial }=f(\frac{1}{m_{initial}}-1)$
$\implies d_{\circ\space initial}=(-34cm)(\frac{1}{\frac{1}{3}}-1)=-68cm$
Similarly $d_{\circ \space fianl }=f(\frac{1}{m_{final}}-1)$
$\implies d_{\circ\space final}=(-34cm)(\frac{1}{\frac{1}{4}}-1)=-102cm$
Now $|d_{\circ\space final}-d_{\circ\space initial}|=|-102-(-68)|=34cm$