Compare a signed dice difference with its absolute value

The source die law is

$E(X)=1(1/6)+2(1/6)+3(1/6)+4(1/6)+5(1/6)+6(1/6)=7/2$. Alternatively symmetry gives $(3+4)/2=7/2$. The Teacher guidance places the main emphasis on the distribution of Y onwards.

$E(X^2)=1^2(1/6)+2^2(1/6)+3^2(1/6)+4^2(1/6)+5^2(1/6)+6^2(1/6)=91/6$. Hence $\operatorname{Var}(X)=E(X^2)-[E(X)]^2=91/6-(7/2)^2=35/12$, as required.

Use the source model of independent fair throws: every ordered pair in the six-by-six table has the same probability $1/36$, as emphasized in the Teacher note. Rows are $X_1$ and columns $X_2$; each cell is $Y=\lvert X_1-X_2\rvert$.

Group the equal cell values:

$E(Y)=0(\frac16)+1(\frac5{18})+2(\frac29)+3(\frac16)+4(\frac19)+5(\frac1{18})=35/18$.

$Z^2=(X_1-X_2)^2$ and $Y^2=\lvert X_1-X_2\rvert^2=(X_1-X_2)^2$. They are equal for every outcome, so $E(Z^2)=E(Y^2)$. The answer is yes.

$E(Z)=E(X_1-X_2)=E(X_1)-E(X_2)=0$, while $E(Y)=E(\lvert X_1-X_2\rvert)>0$. Thus their means differ. Now $\operatorname{Var}(Z)=E(Z^2)-[E(Z)]^2$ and $\operatorname{Var}(Y)=E(Y^2)-[E(Y)]^2$. The second moments are equal by (a), but the subtracted squared means differ. Hence the variances are not equal; in fact $\operatorname{Var}(Z)>\operatorname{Var}(Y)$ (direction made explicit from the source argument).