By current division Rule,

$\bar{I_1}=\dfrac{10\angle0}{10+3-j4}\times25\angle90^0 \\ =\dfrac{10\angle0}{13-j4}\times25\angle90^0$

$\bar{I_1}$ $=\dfrac{(10\times25)\angle90^0}{13.601\angle-17.103} \\ =18.381\angle(90+17.103) \\ =18.381\angle107.103^0$

$\bar{I_2}$ $=\dfrac{(3-j4)}{(10+3-j4)}\times(25\angle90) \\ =\dfrac{(5\angle-53.13)(25\angle90)}{13.601\angle-17.103} \\ =\bigg(\dfrac{5\times25}{13.601}\bigg)\angle(-53.13+90+17.103) \\ =9.191\angle53.973^0$