Answer
(a) $329\Omega $
(b) $794\Omega $
Work Step by Step
(a) We know that
$\epsilon=IR_{eq}$
We plug in the known values to obtain:
$12.0V=(0.075)(R_{eq})$
$\implies R_{eq}=160\Omega $
Now we can find the value of $ R_3$ as
$\frac{1}{160}=\frac{1}{225}+\frac{1}{(225+R_3)}$
$\frac{1}{160}=\frac{450+R_3}{50625+225R_3}$
This simplifies to:
$ R_3=329\Omega $
(b) Since $ R_2$ and $ R_3$ are in series, they share the same voltage.
We know that
$ I_2=\frac{2.65}{225}$
$ I_2=0.0117A $
and
$ I_2R_3=12.0-2.65=9.35$
$\implies R_3=\frac{9.35}{0.0117}=794\Omega $