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I'm given that $X$ and $Y$ are independent and that $\operatorname*{Var}(X)=3$ and $\operatorname*{Var}(Y)=2$, and want to show that $\operatorname*{Var}(X+Y)=\operatorname*{Var}(X)+\operatorname*{Var}(Y)=5$. This only holds if $E[X^2],E[Y^2]<\infty$. Is this the case?

Also what about $\operatorname*{Var}(X-Y)$?

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    Your first question is a little ambiguous. Are you asking whether it is in fact the case that $E\left[X^2\right]$ and $E\left[Y^2\right]$ are finite? Or whether it is the case that $V(X+Y) = V(X) + V(Y)$ only if $E\left[X^2\right]$ and $E\left[Y^2\right]$ are finite? – 2017-01-21
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    If $E[X^2]$ and $E[Y^2]$ are finite. – 2017-01-21
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    You know that the variances are finite. There has to be a formula you know that relates $E\left[X^2\right]$ to $\operatorname{Var} (X)$, no? – 2017-01-21
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    The variance is $V[X]=E([X-E[X])^2$ by definition. and if $E[X^2]$ is finite then $V[X]=E[X^2]-(E[X])^2$ – 2017-01-21

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Note that $Var(X)=E[X^2]-(E[X])^2$. Now if $E[X^2]$ is finite than $Var(X)$ is finite, which is the actual requirement(Var should be finite). So in our case its even better as variance itself is finite so no problem. Now regarding $Var(X-Y)$ note \begin{align*} Var(X-Y)&=E[(X-Y-E[X]+E[Y])^2]=E[((X-EX)-(Y-EY))^2]\\ &=E[(X-EX)^2]+E[(Y-EY)^2]-2E[(X-EX)]E[(Y-EY)]\\ &=VarX+VarY \end{align*} Note that we used independence of $X$ and $Y$ in the second line to say $E[(X-EX)(Y-EY)]=E[X-EX]E[Y-EY]$ and the fact that $E[X-EX]=0$ in the third line. Same proof goes for $Var(X+Y)$.