Let's start with the first covariance:
$$\text{Cov}(A\cdot B,C)=\mathbb{E}[A\cdot B\cdot C]-\mathbb{E}[A\cdot B]\mathbb{E}[C]$$
You know the means of $A,B,C$, namely $\mu_A,\mu_B,\mu_C$, and the pairwise correlation between $A,B$, which I will denote as $\rho_{A,B}$. In case you also have access to the standard deviations of $A,B$, written $\sigma_A,\sigma_B$, you can rewrite
$$\mathbb{E}[A\cdot B]=(\rho_{A,B}\sigma_A\sigma_B+\mu_A\mu_B)$$
Now your desired covariance can be written as:
$$\text{Cov}(A\cdot B,C)=\mathbb{E}[A\cdot B\cdot C]-(\rho_{A,B}\sigma_A\sigma_B+\mu_A\mu_B)\mu_C$$
Similarly, you can write the second covariance as:
$$\text{Cov}(A\cdot B,B\cdot C)=\mathbb{E}[A\cdot B^2\cdot C]-(\rho_{A,B}\sigma_A\sigma_B+\mu_A\mu_B)(\rho_{B,C}\sigma_B\sigma_C+\mu_B\mu_C)$$
In both cases you have a term that cannot be computed with the information at disposal. In particular, note that we cannot say anything about $\mathbb{E}[A\cdot B\cdot C]$ and $\mathbb{E}[A\cdot B^2\cdot C]$. You can take a look at this example in Mathematics Stack Exchange, where the particular case of your problem when $A,B,C$ are pairwise independent is discussed. In general, note that factorizing the joint distribution requires information about the joint behavior of two of the random variables conditioned on the other (e.g. $A,B|C$), which cannot be inferred with your information.
Since you were talking about estimating the autocorrelations, in case you have access to samples of $A,B,C$ (namely $a_i,b_i,c_i$) one option is to use the sample average. That is
$$ \mathbb{E}[A\cdot B\cdot C]\approx \frac{1}{N}\sum_{i=1}^{N}a_i\cdot b_i\cdot c_i$$
You can bound the quality of the estimator above as a function of $N$ using Chebyshev's inequality. Given that you also don't have access to the variance of $A\cdot B\cdot C$ (but you can compute the sample variance) you can take a look at this post.
