$\newcommand{\Cov}{\operatorname{Cov}}$As an extension of this question (covariance of normal distribution), I'd like to give a specific example, and ask what we can deduce about this by imposing assumptions.
Suppose $X\sim N(0,4)$ and $Y\sim N(0,1),$ and let the function $Z=2+3X+Y$.
We can deduce the mean of $Z$ quite easily, and we can find an expression for the variance (in terms of $\Cov(X,Y)$). Is there anything we could say beyond this? I do not think we can make any statements about $\Cov(X,Z)$ or $\Cov(Y,Z)$, without knowing $\Cov(X,Y)$, which we cannot calculate.
Suppose we impose that the vector $(X,Y,Z)$ is multivariate normal. Could we say anything then?
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$\begingroup$Let's use the definitions. Put $Z=2+3X+Y.$ Then $Cov(X,Z)=Cov(X,2+3X+Y)=Cov(X,2)+3Cov(X,X)+Cov(X,Y)=0+3Var(X)+Cov(X,Y).$
In general, just use the linear properties of covariance.
$\endgroup$ $\begingroup$$$\operatorname{cov}(X,Z) = \operatorname{cov}(X,2+3X +Y) = 3\cdot 3 + \operatorname{cov}(X,Y) $$
Suppose we impose that the vector $(X,Y,Z)$ is multivariate normal. Could we say anything then?
If you mean could we say more about variances and covariances of linear functions of $X$ and $Y,$ we cannot, because those are completely determined by the variances and the covariance of $X$ and $Y.$ However, if any nonlinear functions were involved, then we would probably be at a loss without knowing more about the joint distribution of $X$ and $Y$ than their covariance, but if we assumed joint normality, then then more could be said. For example, we would know the distribution of $X^4$ in that case.
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