Yes they are. This is Proposition I.9.2 in Theory of Operator Algebras I by Takesaki.

Proof:

1. => 2): suppose $\pi(a)\xi = 0$ for all $a$. Then $(\pi(a)\eta|\xi) = 0$ for all $\eta\in h$ hence $\xi=0$.
2. => 1): Take $\xi \in h$ orthogonal to all $\pi(a) \eta$. Then from $(\xi| \pi(a^* a) \xi)= 0$ it immediately follows that $\xi =0$.

Yes they are. This is Proposition I.9.2 in Theory of Operator Algebras I by Takesaki.

Short proof:

2. => 1): suppose $\pi(a)\xi = 0$ for all $a$. Then $(\pi(a)\eta|\xi) = 0$ for all $\eta\in h$ and $a\in\mathcal{A}$ hence $\xi=0$.

3. => 2): Take $\xi \in h$ orthogonal to all $\pi(a) \eta$. Then from $(\xi| \pi(a^* a) \xi)= 0$ for all $a$ it follows that $\xi =0$.