Nori's Mixed Motives and Realisation Functors

The conjecture 51 in Levine's Mixed Motives in Handbook of K-theory

http://www.math.illinois.edu/K-theory/handbook/1-429-522.pdf

states that the functor induced by $hs:\text{ECM} \rightarrow \text{MHS}$, denoted by, $$\mathfrak{hs}:\text{NMM}(k)_{\mathbb{Q}} \rightarrow \text{MHS}_{\mathbb{Q}}$$ where $\text{NMM}(k)_{\mathbb{Q}}$ is Nori's mixed motives over a field $k$ (which admits an embedding $\sigma: k \rightarrow \mathbb{C}$) tensored by $\mathbb{Q}$. I am not clear with the construction of the functor $\mathfrak{hs}$ (I guess this is the Hodge realisation functor of Nori's mixed motives), could anyone give me some references? I also have two questions about the properties of $\mathfrak{hs}$.

1. Is $\mathfrak{hs}$ (conjectured to be) exact or not? Has the exactness of it been proved?

2. If $\mathfrak{hs}$ is exact, then it has a derived functor, $$D\mathfrak{hs}:D^b(\text{NMM}(k)_{\mathbb{Q}}) \rightarrow D^b(\text{MHS}_{\mathbb{Q}})$$ In Harrer's thesis,

https://arxiv.org/abs/1609.05516

he constructs a functor,

$$R_n:\text{DM}_{gm}(k,\mathbb{Q}) \rightarrow D^b(\text{NMM}(k)_{\mathbb{Q}})$$ The composition of it with $D\mathfrak{hs}$ is a functor

$$D\mathfrak{hs} \circ R_n :\text{DM}_{gm}(k,\mathbb{Q}) \rightarrow D^b(\text{MHS}_{\mathbb{Q}})$$ Is this functor equivalent to the Hodge realisation functor? Any references?

About 1: since the forgetful functor $MHS_{\mathbb{Q}}\to \mathbb{Q}-Vect$ is exact and faithful, it suffices to verify that the singular cohomology functor from Nori motives is exact. The latter statement should be an immediate consequence of the definition of Nori motives.
• Thank you, I know this book, but I have not read it. I guess 2 follows from 1, as the operator $R_n$ from $\text{DM}_{\text{gm}}(k,\mathbb{Q})$ to $D^b(\text{NMM(k)}_{\mathbb{Q}})$ is compatible with Betti realisation on both categories, while the Betti realisation of $D^b(\text{NMM(k)}_{\mathbb{Q}})$ is the derived functor of $\text{forgetful functor} \circ \mathfrak{hs}$? Thank you very much for you patience with my naive questions! – Wenzhe Jul 8 '17 at 10:34