Assume a random variable $R=T+X$,where T is a Gaussian random variable with $E[T]=0$ and $VAR[T]=9$,and X is independent of T with PDF
$f_x(x)$=$\frac{1}{6}$ for $-3 \le x \le 3$
Find $Cov[T,R]$
I know $Cov[T,R]$=$E[T,R]-E[T]E[R]$,but i don't know how to calculate $E[T,R]$
$\text{Cov}$ is biadditive in the sense that: $$\text{Cov}(U+V,W)=\text{Cov}(U,W)+\text{Cov}(V,W)\tag1$$ and: $\text{Cov}(U,V+W)=\text{Cov}(U,V)+\text{Cov}(U,W)\tag2$
This of course under the condition that the covariances on the RHS exist.
Applying that here we find:$$\text{Cov}(T,R)=\text{Cov}(T,T+X)=\text{Cov}(T,T)+\text{Cov}(T,X)=\text{Var}(T)+0=9$$
The only thing that has to be checked is that $\text{Cov}(X)$ exists, wich is not really difficult.
There is no need for calculation $\mathbb ETR$ (if that's what you mean).
If you insist on doing so then make use of $$\mathbb ETR=\mathbb E(T^2+TX)=\mathbb ET^2+\mathbb ETX=\mathbb ET^2+\mathbb ET\mathbb EX$$
\begin{split} \mathbb{E}\left[ TR\right]=&\mathbb{E}\left[ T(T+X)\right] \\ = & \mathbb{E}\left[ T^2+TX)\right] \\ = & \mathbb{E}\left[ T^2\right] +\mathbb{E}\left[TX\right] \\ = & 9 + \mathbb{E}\left[T\right]\mathbb{E}\left[X\right] \\ = & 9+0 \\ = & 9 \end{split} In the thrid line we used the fact that if $T \sim \mathcal{N}(\mu,\sigma^2)$, then $\mathbb{E}\left[ T^2\right]=\text{Var}(T)-(\mathbb{E}\left[T \right])^2=\sigma^2-\mu^2$, and also the fact that $T$ and $X$ are independent so we can factor the expectation of their product.