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 numerical equivalence class of curves $C$ that contains a curve missing each codimension $1$ subset. The for $\mathcal O(D)$ to have a nonzero global section, it must certainly have a nonzero global section restricted to some curve in $C$, 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 we can choose $f= C \cdot D$ and $-C \cdot D$ and $c=0$ and $- C \cdot K$ respectively.
To construct such a $C$, if the variety is projective, we can take the intersection with a linear subspace of the appropriate dimension and get a curve we can move around anywhere. If it is not then I think a birational transformation to a projective variety will work, but I'm not sure.
No idea about question $3$.