Evaluating a matrix element of a $3\times 3$ Hamiltonian in terms of Gell-Mann matrices A generic $3\times 3$ Hamiltonian can be expressed in terms of eight Gell-Mann matrices ($\lambda$) as
\begin{align}
{\cal H} &= h_{0} I +  H= h_{0} I + \sum_{\alpha=1}^{8} h_{\alpha} \lambda_{\alpha}, \\
{\cal H}|n \rangle &=( h_{0} + \epsilon_{n}) | n \rangle \qquad   \text{For  } n\in \{1,2,3 \},
\end{align}
where I is the identity matrix, $|n \rangle $ are eigenvectors and $\epsilon_{n}$ denote eigenvalues of $H$. This decomposition enable one to determine different qunatities, e.g., Berry curvature, in terms of matrix elements of Gell-Mann matrices.
Using properties of $\lambda$ matrices, we can evaluate the non-diagonal matrix elements of $\langle m |[\lambda_{\alpha}, H]| n \rangle $,
\begin{align}
\lambda_{\alpha}^{mn} = 2 {\rm i} \sum_{\gamma, \beta} f_{\alpha \beta \gamma} h_{\beta} \frac{ \lambda_{\gamma}^{mn}}{\epsilon_{m} -\epsilon_{n}} =  \sum_{\gamma} {\cal F}_{\alpha \gamma}^{mn}   \lambda_{\gamma}^{mn},
\end{align}
where $f_{\alpha \beta \gamma}$ are SU(3) structure constants and we have employed $[\lambda_{\alpha}, \lambda_{\beta}] = {\rm i} \sum_{\gamma} f_{\alpha \beta \gamma}\lambda_{\gamma} $.
My question is whether it is possible to evaluate
\begin{align}
 \frac{  \lambda_{\alpha}^{mn} }{\epsilon_{m} -\epsilon_{n}} = \sum_{\gamma} {\cal K}_{\alpha \gamma}^{mn}   \lambda_{\gamma}^{mn},
\end{align}
such that $\cal K$ can expressed merely in terms of $h$ and matrix elements of $\lambda$?
 A: You might be  trapping yourself in your notation. First, appreciate that $[\lambda_\alpha,{\cal H}]= [\lambda_\alpha,  H ]$, so w.l.o.g. set $h_0=0$. Moreover, since $H|n\rangle=\epsilon_n|n\rangle$, the $\epsilon_n$s are functionals of h.
You are considering
$$
[\lambda_{\alpha}, H]= 2i \sum_{\beta, \gamma} f^{\alpha \beta\gamma}h_\beta \lambda_\gamma.
$$
Defining $\Lambda ^\alpha (m,n)=\langle m |\lambda_{\alpha} | n \rangle $, the matrix elements of the above linear equation are
$$
\lambda_{\alpha}(m,n) \equiv \langle m |[\lambda_{\alpha},H] | n \rangle =(\epsilon_n-\epsilon_m) \Lambda^\alpha (m,n)\\  = 2 {\rm i} \sum_{\gamma, \beta} f_{\alpha \beta \gamma} h_{\beta}   \Lambda_{\gamma}(m,n)  \equiv   \sum_{\gamma} (\epsilon_n-\epsilon_m){\cal F}_{\alpha \gamma}(mn)   \Lambda_{\gamma}(m,n).
 $$
(NB. You have interchanged n with m in your eigenvalues' expression, not clear why.)
For off diagonal elements, recalling  (m,n) are just indices never summed over, you have the eigenvalue-1 equation
$$
\Lambda_{\alpha} (m,n)  = \sum_{\gamma} {\cal F}_{\alpha \gamma} (m,n)    \Lambda_{\gamma} (m,n) .
 $$
Evidently ${\cal, F, K,...}$, etc, are all functionals of h, and hence the corresponding eigenvalues $\epsilon_m$.
It would be easiest for you to simply choose $h_1 =1$ and the rest vanishing, to check your expressions. E.g., $\vec \Lambda (1,-1)= (0,i,1,0,0,0,0,0)^T$, with $\epsilon_{-1}-\epsilon_{1}=-2$, hence ${\cal F}_{\alpha \gamma}(1,-1)=-i f_{\alpha  1\gamma}$, imaginary antisymmetric, etc... Recall $1=f_{ 3  1   2}=2f_{ 7 1 4}=2f_{ 5 1 6}$. So ${\cal F}_{ 23}(1,-1)=i$.  Observe the evident eigenvector with eigenvalue 1. You might as well have found it this way, if you did not have it!
But I still don't fully  fathom your objective. If you were interested in exponentiation  of H, you might well consider this.
