For phasors, \$I=I_{_0}\cos\left(\phi\right)\$. But the problem used sine, not cosine. I would first convert to cosine. For that, subtract \$90^\circ\$ to get to a cosine and because (for power reasoning) current is conjugate to voltage, it is \$180^\circ\$ out of phase. So \$\phi=120^\circ-90^\circ\pm 180^\circ\$ (whichever is less than \$180^\circ\$.) In this case, \$\phi=-150^\circ\$. So \$I=2\cdot\left[\cos\left(-150^\circ\right)+j\sin\left(-150^\circ\right)\right]=-\sqrt{3}-j\$. Which appears to be what the solution says.
As far as the admittance goes, it really is just the reciprocal of impedance. And you were right to calculate impedance the way you did. I agree with it.
Here's what my calculator said (after I set \$\omega=2\$) when I typed in 1/(((3|2)+1F)|3F):
$$0.710059172 + j6.29585799, 6.33577239\: \angle\: 83.5652608^\circ$$
So I think you are fine with your approach. You know how to compute the impedance. And the admittance is just the reciprocal of it. That's all.
To turn \$\frac{c}{a+bi}\$ into \$a^{'}+b^{'}i\$ just follows a process:\$\frac{c}{a+bi}\cdot\frac{a-bi}{a-bi}=c\frac{a}{a^2+b^2}-c\frac{b}{a^2+b^2}i\$. In this case, the impedance you should compute from ((3|2)+1F)|3F is \$0.0176886792 - j0.156839623\$. So the admittance is then just \$\frac{0.0176886792}{0.0176886792^2+0.156839623^2}-\frac{0.156839623}{0.0176886792^2+0.156839623^2}i\$. And that does get you the right result.