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
Since the die is fair, there are $6$ equally likely outcomes for each of the five rolls, so there are $6^5$ possible outcomes for the five rolls.
We have a sequence of five outcomes. For the favorable cases, choose two of the five positions in the sequence for the ones and two of the remaining three positions in the sequence for the threes. The final open position in the sequence must be filled with a six. Therefore, there are $$\binom{5}{2}\binom{3}{2} = \frac{5!}{3!2!} \cdot \frac{3!}{2!1!} = \frac{5!}{2!2!1!}$$ favorable sequences.
Hence, the probability that two ones, two threes, and a six are obtained when a fair die is rolled five times is $$\frac{\dbinom{5}{2}\dbinom{3}{2}}{6^5} = \left(\frac{5!}{2!2!1!}\right)\left(\frac{1}{6^5}\right)$$