With the sample space $\Omega=\left\{1,2,3,4,5,6\right\}^2=\left\{\left(1,1\right),\left(1,2\right),\dotsc,\left(6,5\right),\left(6,6\right)\right\}$, we can easily find what is being asked.
The range of a Random Variable is the set of values it can assume, so:
$\mathrm{Range}\left(X\right)=\left\{2,3,\dotsc,11,12\right\}$ as these are the possible sums of the sides of a fair die rolled twice.
$\mathrm{Range}\left(Y\right)=\left\{1,\dotsc,6\right\}$ as whatever is rolled, any of these values could be the maximum.
$\mathrm{Range}\left(Z\right)=\left\{0,1,\dotsc,5\right\}$ as these are the possible differences between the values of the two rolls, note that we have $0$ when $d_1=d_2$ and 6 isn't possible as the largest difference will occur when we get $\left(6,1\right)$ or $\left(1,6\right)$.
$\mathrm{Range}\left(W\right)$ is a little more difficult as we need to calculate $W\left(\omega\right)$, $\forall\omega\in\Omega$. When done, we find that $\mathrm{Range}\left(W\right)=\left\{0,3,5,7,8,9,11,12,15,16,20,21,24,27,32,35\right\}$
A partition of a random variable is the set of points $\omega\in\Omega$ that give rise to each possible value that the random variable can take.
The partitions $A_X$ and $A_Z$ can now be found
\begin{align*}
A_{X_1=2} &= \left\{\left(1,1\right)\right\} \\
A_{X_2=3} &= \left\{\left(1,2\right),\left(2,1\right)\right\} \\
A_{X_2=4} &= \left\{\left(1,3\right),\left(2,2\right),\left(3,1\right)\right\} \\
&\vdots \\
A_{X_{10}=11} &= \left\{\left(5,6\right),\left(6,5\right)\right\} \\
A_{X_{11}=12} &= \left\{\left(6,6\right)\right\} \\
\end{align*}
and
\begin{align*}
A_{Z_1=0} &= \left\{\left(1,1\right),\left(2,2\right),\dotsc,\left(5,5\right),\left(6,6\right)\right\} \\
A_{Z_2=1} &= \left\{\left(1,2\right),\left(2,3\right),\dotsc,\left(3,2\right),\left(2,1\right)\right\} \\
&\vdots \\
A_{Z_6=5} &= \left\{\left(1,6\right),\left(6,1\right)\right\} \\
\end{align*}
In response to the comment about why the value $5\times12\not\in\mathrm{Range}\left(W\right)$, we see that $W\left(\omega\right)=X\left(\omega\right)\times Z\left(\omega\right)$, so therefore it isn't possible to obtain $60=12\times5$ because the only value $\omega\in\Omega$ such that $X\left(\omega\right)=12$ is $\omega=\left(6,6\right)$ but clearly $Z\left(\omega\right)=0$ if $\omega=\left(6,6\right)$. Similarly for $Z\left(\omega\right)=5$, we need $\omega=\left(1,6\right)$ or $\omega=\left(6,1\right)$ and then clearly $X\left(\omega\right)=7$ for both of these $\omega$.
In order to compute $\mathrm{Range}\left(W\right)$, it is necessary to check the value of $W\left(\omega\right)=X\left(\omega\right)\times Z\left(\omega\right)$ for all $\omega\in\Omega$ where $\Omega$ is defined as above.
Best Answer
What you need are the respective frequencies.
On a total of $36$ possibilities, for the maximum,
$$\begin{matrix}1&2&3&4&5&6\\\hline1&3&5&7&9&11\end {matrix},$$
and for the minimum,
$$\begin{matrix}1&2&3&4&5&6\\\hline11&9&7&5&3&1\end {matrix}.$$
The rest is yours.