First let me give some cheap examples of CY manifolds that satisfy condition of 1) while they are not tori quotients.
Example. Let $X$ be any CY manifold with $h^{1,1}(X)=1$, for example $X$ can be a quintic in $\mathbb CP^4$. Then the conclusion of 1) holds, because all Kahler Ricci flat metric on $X$ are proportional. Now, to get an example with $h^{1,1}(X)=2$ one can just take a direct product $X_1\times X_2$ of CY manifolds satisfying $h^{1,1}(X_i)=1$.
So the list of manifolds with $c_1=0$ satisfying 2) is larger than tori quotients. But maybe all these examples can be understood. For example, it should be easy to show that a $K3$ surface does not satisfy 1). I'll give a brief proof in the case of a generic K3
that does not have $-2$ curves.
EDDITED. I put more details here so that it becomes clear that was said previously is correct.
The good thing about harmonic forms is that the sum of two harmonic forms is harmonic.
By Nakai-Moishenzon (for Kahler surfaces), the Kahler cone
of a K3 without $-2$ curves coincides with a connected component
of $(1,1)$ classes with positive square in $H^{1,1}$.
Now, by Yau in any such class there is a unique Ricci-flat metric. And also obviously
there exists a unique harmonic metric. Suppose that assumption 1) holds for our K3.
Then it is obvious that this means that each harmonic form $w$ with $\int_{K3}w^2>0$
in the correct component of the cone
is Ricci flat. Let me get a contradiction from this
So let $w_1$ and $w_2$ be two Kahler Ricci-flat forms harmonic with respect to the metric defined by $w$ on a $K3$ surface, and moreover that $aw_1+bw_2$ is Ricci flat again. Then we know that $(aw_1+aw_2)^2$ is proportional to $\Omega\wedge \bar\Omega$, where $\Omega$ is the complex volume form on $K3$. Consider now the family of forms $w_1-tw_2$, $t>0$. Take the first moment $t_0$ when we have $\int_{K3}(w_1-t_0w_2)^2=0$. But since $(w_1-tw_2)^2=c_t\Omega \wedge \bar\Omega$ for some $c_t>0$ for all $t$ less than $t_0$ (since this form is Ricci flat)
we conclude that $(w_1-t_0w_2)^2$ is equal to zero point-wise. So the kernel of $w_1-tw_2$ should define a holomorphic folitation on $K3$. Since for $K3$ it holds $h^{1,1}=20$, we should get a tremendous amount of holomorphic foliations on it, but this is clearly impossible.
This is still not a 100% complete reasoning, but I am sure it can be completed. So it seems to me that the complete list of manifolds satisfying 1) are all manifolds that are finite quorients of a torus times a collection of CY manfiolds with $h^{1,1}=1$. The idea is simple: whenever you have two Ricci-flat forms such that $w_1+tw_2$ is Ricci flat for small $t$, this should lead to a local metric splitting of the manifold into direct product. Then we should just use De-Rahm decomposition theorem.
On a $n$-dimensional Kähler manifold $(X,\omega)$, the Ricci form is (minus) the curvature of the canonical bundle $K_X$ endowed with the induced metric. Thus, if $X$ has zero Ricci curvature then its canonical bundle is flat. Thus, the structure group can be reduced to a subgroup of the special linear group $SL(n,\mathbb C)$.
However, Kähler manifolds already possess holonomy in $U(n)$, and so the (restricted) holonomy of a Ricci flat Kähler manifold is contained in $SU(n)$. Conversely, if the (restricted) holonomy of a $2n$-dimensional Riemannian manifold is contained in $SU(n)$, then the manifold is a Ricci-flat Kähler manifold.
In the case when $X$ is compact the celebrated solution of Yau to the Calabi problem asserts that if $c_1(X)=0$ then $X$ posses a metric with vanishing Ricci curvature. For the non compact case, there are some (among others) results by Tian and Yau which concerns the existence of complete Ricci-flat Kähler metrics on quasiprojective varieties. One of their main theorems is the following:
Suppose that $X$ is a smooth complex projective variety with ample anticanonical line bundle (i.e. a Fano manifold), and that $D\subset X$ is a smooth anticanonical divisor. Then $X\setminus D$ admits a complete Ricci-flat Kähler metric.
Best Answer
The whole point is that if the metric is Kähler, then the Chern connection coincides with the complexification of Levi-Civita connection of the riemanian metric on the underlying real manifold given by the real part of the Kähler hermitian metric.
Moreover, this happens if and only if the starting metric is Kähler (and is indeed my favorite definition of Kähler metric).
So, let me give you a hint.
If $\omega$ is any hermitian metric on $X$, then $\det\omega$ is the induced metric on $\det T_X$ (this line bundle is usually referred as to the anticanonical bundle $K_X^{*}$). The Chern curvature $\Theta$ of this metric is given by $$ i\Theta(K_X^*,\det\omega)=-i\partial\bar\partial\log\det\omega. $$ Now, in general, the Chern curvature (with respect to the induced hermitian metric) of the determinant $\det E$ of a hermitian holomorphic vector bundle $(E,h)$ is given by the trace of the Chern curvature of $E$: $$ i\Theta(\det E,\det h)=i\operatorname{Tr}\Theta(E,h). $$ Therefore, $$ -i\partial\bar\partial\log\det\omega=i\operatorname{Tr}\Theta(T_X,\omega). $$ But $\Theta(T_X,\omega)$ is the complexification of the curvature $R^{LC}$ of the Levi-Civita connection of $X$ $\nabla^{LC}_{X,\Re\omega}$ with respect to $\Re\omega$, where $\Re\omega$ is the induced riemannian metric on the underlying real manifold.
Finally, by definition, the Ricci curvature of $(X,\Re\omega)$ is the trace of $R^{LC}$.
There is a small missing point in my sketchy argument: the riemannian Ricci curvature is a symmetric $2$-controvariant tensor, while the Ricci form is a closed real $(1,1)$-form. You have to show that in fact the riemannian Ricci curvature of (the real part of) a Kähler metric is invariant with respect to the complex structure $J$, then you apply the $1-1$ correspondence between real $(1,1)$-forms and symmetric $J$-invariant $2$-controvariant tensors.