Properties of Covariances When Adding and Subtracting Suppose $T = Y_{1} + Y_{2} + Y_{3}$ where $Y_{i}$ is a random variable. Let $U = Y_{1} + Y_{3} - Y_{4}$. 
$T$ and $U$ are not independent.
How could I make a formula to find Covariance$[T,U]$ if Cov$[Y_{3}, Y_{i}] = p$, and the rest of the $Y_{i}$'s are independent?
 A: Yes, you can, mostly. There is a theorem, Theorem 5.12 in Wackerly, Mendenhall, and Scheaffer, 5th Ed., p. 228, that goes like this:

Let $Y_1,Y_2,\dots,Y_n$ and $X_1,X_2,\dots,X_m$ be random variables. Define 
  $$U_1=\sum_{i=1}^n a_iY_i\qquad\text{and}\qquad U_2=\sum_{j=1}^mb_jX_j$$
  for constants $a_1,a_2,\dots,a_n$ and $b_1,b_2,\dots,b_n.$ Then
  $$\operatorname{Cov}(U_1,U_2)=\sum_{i=1}^n\sum_{j=1}^m a_ib_j\operatorname{Cov}(Y_i,X_j).$$

Before moving on, note that $\operatorname{Cov}(Y,Y)=E\big(Y^2\big)-(E(Y))^2=\operatorname{Var}(Y),$ and that $\operatorname{Cov}(X,Y)=\operatorname{Cov}(Y,X).$
Applying this to your problem, we can say that
\begin{align*}
\operatorname{Cov}(T,U)
&=\operatorname{Cov}(Y_1,Y_1)+\operatorname{Cov}(Y_1,Y_3)-\operatorname{Cov}(Y_1,Y_4)\\
&\quad+\operatorname{Cov}(Y_2,Y_1)+\operatorname{Cov}(Y_2,Y_3)-\operatorname{Cov}(Y_2,Y_4)\\
&\quad+\operatorname{Cov}(Y_3,Y_1)+\operatorname{Cov}(Y_3,Y_3)-\operatorname{Cov}(Y_3,Y_4)\\
&=\operatorname{Var}(Y_1)+p-0+0+p-0+p+\operatorname{Var}(Y_3)-p\\
&=\operatorname{Var}(Y_1)+\operatorname{Var}(Y_3)+2p.
\end{align*}
Without knowing the variances of $Y_1$ and $Y_3,$ this is as far as you can get, unless your expression $\operatorname{Cov}(Y_3,Y_i)=p$ admits $i=3,$ in which case you could obtain
$$\operatorname{Var}(Y_1)+3p.$$
