$$Z=3 X-2 Y \text{, so } E(Z)=3 E(X)-2 E(Y) \text {. }
$$
For two fair dice:
$$
\begin{aligned}
& P(\max =k)=(k / 6)^2-((k-1) / 6)^2=(2 k-1) / 36 \\\\
& E(X)=\Sigma k \cdot(2 k-1) / 36=161 / 36
\end{aligned}
$$
If the two die outcomes are $A$ and $B$, then by definition
$$
\max (A, B)+\min (A, B)=A+B
$$
$$
\begin{aligned}
& X+Y=A+B \\
& \\
& \mathbb{E}[X]+\mathbb{E}[Y]=\mathbb{E}[A]+\mathbb{E}[B]=3.5+3.5=7 .
\end{aligned}
$$
Sum of the two dice has expectation $7$, so $\mathrm{E}(\mathrm{X})+\mathrm{E}(\mathrm{Y})=7$.
$$
\begin{aligned}
& E(Y)=7-161 / 36=91 / 36 \\\\
& E(Z)=3 \cdot(161 / 36)-2 \cdot(91 / 36)=301 / 36 \\\\
& E(Z)=8.36
\end{aligned}
$$
