If you look at the second question, the OP is asking when the cohomology ring can be generated by some *set* of Kähler forms, not by a *single* Kähler form. Thus, while the variety $F$ of full flags in $\mathbb{C}^3$ has two generators, say $\omega_1$ and $\omega_2$ that are *not* Kähler forms, the forms $\eta_1 = 2\ \omega_1+\omega_2$ and $\eta_2=\omega_1+2\ \omega_2$ are both positive and hence Kähler forms. They generate the cohomology ring over the rationals.

In general, if the space is of the form $X = G/P$ where $G$ is complex semi-simple and $P$ is a minimal parabolic, then $X = U/T$ where $U$ is a compact form of $G$ and $T = U\cap P$ is a maximal torus in $U$. The cohomology ring of $X$ will be generated in degree $2$ by a vector space $V$ of $U$-invariant $(1,1)$-forms, and one can find, in the positive cone in this vector space, a basis of generators of the cohomology ring of $X$ (over the rationals), and these generators will all be Kähler forms.

Thus, for example, when $F$ is the full flag variety of $\mathbb{C}^n$, $G=\mathrm{SL}(n,\mathbb{C})$, $P$ is the upper triangular elements in $G$, $U=\mathrm{SU}(n)$, and $K=\mathbb{T}^{n-1}$ is the maximal torus of diagonal elements in $G$. The vector space $V$ has dimension $n{-}1$ and there is a positive 'orthant' in $V$ that consists of Kähler forms that are $\mathrm{SU}(n)$-invariant. These suffice to generate (rationally) the cohomology of $F$ (with, of course, some relations).

Obviously, the case of minimal parabolics is not the only case in which this happens, but I'd have to think about this more to 'remember' the criterion, which I think I did know, at one time.