Let $X, Y$ be random variables. Then:
$1.$ If $X, Y$ are independent and $Z=X*Y$, then $Cov(X,Z)=EY*VarX$
$2.$ If $X, Y$ are independent and $Z=X*Y$, then $Cov(X,Z)=EY*VarX+VarY$
$3.\ Var(X+Y)=VarX+VarY-2Cov(X,Y) $
$4.$ If $X, Y$ are independent, then $Var(X*Y)=EX^2*VarY+(EY)^2*VarX$
$5.$ If $X, Y$ are independent, then $Var(X+Y)>Var(X-Y)$
My question is: is there any scheme for doing such exercises and do I have to know any specific formulas (except that $VarX=EX^2-(EX)^2$ and $Cov(X,Y)=E(XY)-EXEY$) to make these? I will appreciate any effort at helping me with this problem
edit 2:
$ad.1.$ $Cov(X,Z)=Cov(X,XY)=E(XXY)-E(X)E(XY)=E(X^2Y)-E(X)E(X)E(Y)=E(X^2)E(Y)-(EX)^2E(Y)=E(Y)[E(X^2)-(EX)^2] =EY*VarX $
$ad.2.$ $Cov(X,Z)=Cov(X,XY)=E(XXY)-E(X)E(XY)=E(X^2Y)-E(X)E(X)E(Y)=E(X^2)E(Y)-(EX)^2E(Y)=E(Y)[E(X^2)-(EX)^2] \neq EY*VarX + VarY$
$ad.3.$ $Var(X+Y)=VarX+VarY+2Cov(X,Y) \neq VarX+VarY-2Cov(X,Y) $
$ad.5.$ $Var(X+Y)=Var(X-Y)$, so there's no inequality
These seem like introductory exercices on the basis definitions. Like you've used, if $X, Y$ are random variables $$ \mathrm{Cov}(X,Y) = \mathrm{E}(XY) - \mathrm{E}(X)\mathrm{E}(Y), $$ and $\mathrm{Var}(X) = \mathrm{Cov}(X,X)$.
Your solution to the first is not right though. $$ \mathrm{E}(X^2) \not = \mathrm{E}(X)^2 $$
It might help to think about a distribution for this. Let $X\sim N(0,1)$ then $\mathrm{E}(X) = 0$. However $\mathrm{E}(X^2) \not = \mathrm{E}(X)^2 = 0$. This makes sense because $X^2$ has nothing but positive values. Even more, the distribution is $X^2 \sim \chi^2$. So $\mathrm{E}(X^2) = 1$. (Or notice how here $\mathrm{Var}(X) = \mathrm{E}(X^2) - \mathrm{E}(X)^2$ which implies $1 = \mathrm{E}(X^2) - 0$)