I am not familiar with triangulated categories so these questions might be too basic (but I did not find any answers by google). Also, the question can be formulated in purely triangulated category setting, but I used derived category anyway...
For any smooth variety $X$ over $\mathbb C$, we denote by $D(X)=D(Coh(X))$ its (bounded) derived category of coherent sheaves on $X$. Let $H^0$ be the cohomology functor $D(X)\to Coh(X)$ induced from the standard t-structure, and also $H^* :D(X) \to D(X)$ by direct sum of all the shifted cohomology. Let $F$ be any triangulated functor $D(X)\to D(Y)$. I am thinking about the following (maybe too standard) questions:
Question 0. In some literature, a triangulated functor is also called an exact functor. I am very confused about this name, as it suggests that it should preserve some t-structures, but I did not see any. Is there a reason why we call it exact? (Edit: I confused myself with the notions. Exact is not the same as t-exact)
Question 1. Does $F$ commute with $H^*$?
Question 2. If $E$ splits (i.e. isomorphic to the direct sum of its shifted cohomology), does $F(E)$ split as well?
I guess both questions have negative answers but I cannot give an example. For example, for the last question, I try to consider the example where $E\in Coh(X)$ and $F=Rf_*$ for some $f:X \to Y$. If $Y$ is a point or $f$ is affine, $R f_* E$ will split for trivial reasons. And I think for other general case it will not split, but I don't know how to show it...
Thanks in advance for any comments!