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.
Yes, if you know the joint probability distribution of $X,Y$ then the distributions of $U$ and $V$ are, respectively:
$$\begin{align}
\mathsf P(U=u) & = \sum_\overbrace{x\in\Omega_X, u-x\in\Omega_Y} \mathsf P(X=x, Y=u-x)
\\
\mathsf P(V=v) & = \sum_\overbrace{x\in\Omega_X, v/x\in\Omega_Y} \mathsf P(X=x, Y=v/x)
\end{align}$$
The joint distribution on $U,V$, would similarly be $$\mathsf P(U=u, V=v) = \sum_\overbrace{x\in\Omega_X, u-x\in\Omega_Y, v/x\in\Omega_Y} \mathsf P(X=x, Y=u-x, Y=v/x)$$
However, for any given pair of $U=u,V=v$ there is in fact only two corresponding pair of $X=\Box,Y=\Box$.
$\begin{align}
(U=X+Y) \wedge (V= XY) & \iff
\left(X= \dfrac{U\pm\sqrt{U^2-4V}}{2}\right)
\wedge \left(Y=\dfrac{U\mp\sqrt{U^2-4V}}{2}\right)
\\[4ex]
\therefore \mathsf P(U=u, V=v)
& = \mathsf P\left(X= \dfrac{u\pm\sqrt{u^2-4v}}{2}, Y=\dfrac{u\mp\sqrt{u^2-4v}}{2}\right)
\\[1ex]
& = {\mathsf P\left(X= \tfrac{u+\sqrt{u^2-4v}}{2}, Y=\tfrac{u-\sqrt{u^2-4v}}{2}\right)
+ \mathsf P\left(X= \tfrac{u-\sqrt{u^2-4v}}{2}, Y=\tfrac{u+\sqrt{u^2-4v}}{2}\right)}
\end{align}$
Best Answer
Here is a tabular representation of your joint pmf
I hope this will clarify the situation
Example: in the cell $(X;Y)=(1;2)$ that means: minimun die's value=1 and maximum die's value=2
That is $(D_1;D_2)=(1;2)$ or $(D_1;D_2)=(2;1)$