Use superposition to find voltage V1
(a) The circuit across the active current source is a parallel circuit of C and a series circuit of R and L.


Note that components values are given as admittances (not as impedances).So you can't calculate \$V = IZ\$ (\$Z=\$impedance) butyou have to use \$V = \frac{I}{Y}\$ (\$Y=\$admittance).
The total admittance is \$Y=Y_C + Y_{RL}\$ (parallel admittance are calculated like series impedances or resistances: just add them) and
\$Y_{RL}=\frac{Y_R Y_{L}}{Y_R + Y_{L}}\$ (series admittance are calculated like parallel impedances or resistances: product divided by sum) 


If you follow the steps you get exactly the given solution.

(b) Now the circuit across the active current source is a parallel circuit of L and a series circuit of R and C.

You get V1 by 


using the current divider formula to get \$I_L\$ and \$I_{RC}\$
\$V1 = \frac{I_C}{Y_C}\$ where \$I_C = I_{RC}\$ (because R and C are in series).