Answer
(a) $29V$
(b) $I_{65}=0.45A$
$I_{25}=1.2A$
$I_{170}=0.17A$
Work Step by Step
We can find the equivalent resistance as
$\frac{1}{R_{eq}}=\frac{1}{65}+\frac{1}{25}+\frac{1}{170}$
$\implies R_{eq}=16.322\Omega$
Now we can find the emf as follows:
$V=IR_{eq}$
We plug in the known values to obtain:
$V=(1.8)(16.322)$
$V=29V$
(b) We can find the current passing through each resistor as
$I_{65}=\frac{V}{R}=\frac{29.4}{65}=0.45A$
$I_{25}=\frac{V}{R}=\frac{29.4}{25}=1.2A$
$I_{170}=total\space current -I{65}-I_{25}=1.8-1.176-0.452=0.17A$
