De-equivariantization by Rep(G)

I'm trying to understand Proposition 2.9 of this paper on weakly group theoretical fusion categories.

First of all I have a problem with understanding the settings for de-equivariantization process. It is written on page 5 of the paper that one needs $\mathcal{E}=\mathrm{Rep}(G)\subset Z(\mathcal{C})$ such that $\mathcal{E}$ embeds into $\mathcal{C}$ via the forgetful functor $Z(\mathcal{C})\rightarrow \mathcal{C}$.

Question 1. I would like to know what "embeds" means exactly.

Rephrasing when a functor is an embedding functor is? Is it enough to be injective on objects and to send simple objects into simple objects. Obviously this is necessary.
Question 2. In a concrete example where $\mathcal{C}=\mathrm{Rep}(A)$ for a Hopf algebra $A$ what would be the conditions to be imposed for the corresponding composition to be an embedding.

Is the corresponding category $\mathcal{C}_G$ the category of representations of a Hopf algebra, i.e does it posses a fiber functor?

Question 3. The third question I have is regarding the de-equivariantization $\mathcal{E}'_G$ from the second part of the proof proposition.

I understood the construction of $\mathcal{E}$ and clearly $\mathcal{E}\subset \mathcal{E}'$ since $\mathcal{E}$ is symmetric. The question I have is why after composing to the restriction functor $Z(\mathcal{E}') \rightarrow \mathcal{E}'$ one still has an inclusion.

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Question 1. "embeds" means that the functor ${\mathcal E}\to {\mathcal C}$ is fully faithful. Equivalently, this functor sends simple objects to simple ones and non-isomorphic simple objects to non-isomorphic ones (so your guess is correct).
Question 2: In the situation when ${\mathcal C}=Rep(A)$ this means that the group algebra of $G$ is a Hopf quotient of the Hopf algebra $A$ and the functor ${\mathcal E}\to {\mathcal C}$ is isomorphic to the pullback functor $Rep(G)\to Rep(A)$. We do not expect that the category ${\mathcal C}_G$ has a fiber functor in this situation (It should not be too difficult to construct an explicit counterexample but I do not remember it right now).
Question 3: Here you compose embedding ${\mathcal E}\to {\mathcal E}'$ with functor ${\mathcal E}'\to Z({\mathcal E}')$ (which comes from the braiding on ${\mathcal E}'$) and functor $Z({\mathcal E}')\to {\mathcal E}'$. But notice that the composition ${\mathcal E}'\to Z({\mathcal E}')\to {\mathcal E}'$ is just the identity functor, so in the end of the day you compose embedding ${\mathcal E}\to {\mathcal E}'$ with the identity functor.