The answer to question $1$ is no. For a product of two curves $C_1 \times C_2$ of positive genus, there are infinitely many numerical equivalence classes where $H^0(X,L)$ is not determined by the numerical equivalence class, since for any very ample line bundle $L_1$ on $C_1$ and degree $0$ nontrivial line bundle $L_2$ on $C_2$, $H^0(X,L_1^{\otimes n} \otimes L_2)=0<H^0(X,L_1^{\otimes n} \otimes \mathcal O_X)$, but the two bundles are numerically equivalent, so there are infinitely many numerical equivalence classes without this property.
I would expect that instead you can make the statement: for any numerical equivalence class $D_1$ and ample numerical equivalence class $D_2$, for sufficiently large $n$, $D_1+nD_2$ and $D_1-nD_2$ have this property, by a vanishing of cohomology argument.
Question 2: I think $B_0$ and $B_d$ will look, at least much more like a cone than the middle $B_i$. Consider $\mathbb P^1 \times \mathbb P^1$. $B_0$ and $B_2$ are cones, but $B_1$ is not.
For $B_0$ and $B_d$, choose any curve $C$. The for $\mathcal O(D)$ to have a global section, it must certainly have a global section restricted to $C$, so $C \cdot D\geq 0$. By Serre duality, for $\mathcal O(D)$ to lie in $B_d$, we must have $C \cdot D \leq C \cdot K$. So the answer to the specific question is "yes" for $B_0$ and $B_d$.
No idea about question $3$.