= Solution
Use the usual assumption that the two die scores are <independent random variables>. A fair score has <expected value> $7/2$ and <variance> $35/12$. <Linearity of expectation> and the <variance of a sum> therefore give
$$
\boxed{\mathbb EX=7,\quad\mathbb EY=0,\quad
\operatorname{Var}X=\operatorname{Var}Y=\frac{35}{6}.}
$$
Count the ordered pairs of scores to obtain the <probability mass functions>
$$
P(X=x)=\frac{6-|x-7|}{36},\quad x=2,\ldots,12;\qquad
P(Y=y)=\frac{6-|y|}{36},\quad y=-5,\ldots,5.
$$
On these attainable <probability supports>, \b[the maxima occur at $x=7$ and $y=0$, each with probability $1/6$; the minima occur at $x=2,12$ and $y=-5,5$, each with probability $1/36$]. Outside the respective <probability support>, the mass is zero.
They are not independent: $Y=0$ implies equal die scores and hence $X$ is an <even number>, so $P(X=7,Y=0)=0$, whereas $P(X=7)P(Y=0)=1/36>0$. In fact $\operatorname{Cov}(X,Y)=\operatorname{Var}(S_1)-\operatorname{Var}(S_2)=0$: this is an example of <uncorrelated random variables> that are dependent.
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