Supposing that the fair die has $n$ sides and is rolled $m$ times we
get for the probability of $k$ distinct outcomes by first principles
the closed form
$$\frac{1}{n^m} \times
m! [z^m] {n\choose k} (\exp(z)-1)^k.$$
Let us verify that this is a probability distribution. We obtain
$$\frac{1}{n^m} \times
m! [z^m]
\sum_{k=0}^n {n\choose k} (\exp(z)-1)^k.$$
Here we have included the value for $k=0$ because it is a constant
with respect to $z$ and hence does not contribute to $[z^m]$ where
$m\ge 1.$ Continuing we find
$$\frac{1}{n^m} \times
m! [z^m] \exp(nz) = 1$$
and the sanity check goes through. Now for the expectation we get
$$\mathrm{E}[N] =
\frac{1}{n^m} \times
m! [z^m]
\sum_{k=1}^n k {n\choose k} (\exp(z)-1)^k
\\ = \frac{n}{n^m} \times
m! [z^m]
\sum_{k=1}^n {n-1\choose k-1} (\exp(z)-1)^k
\\ = \frac{n}{n^m} \times
m! [z^m] (\exp(z)-1)
\sum_{k=0}^{n-1} {n-1\choose k} (\exp(z)-1)^k
\\ = \frac{n}{n^m} \times
m! [z^m] (\exp(z)-1) \exp((n-1)z).$$
This is
$$\frac{n}{n^m} \times
(n^m - (n-1)^m) = n \left(1-\left(1-\frac{1}{n}\right)^m\right).$$
This goes to $n$ in $m$ as it ought to. For the next factorial moment
we obtain
$$\mathrm{E}[N(N-1)] =
\frac{1}{n^m} \times
m! [z^m]
\sum_{k=2}^n k (k-1) {n\choose k} (\exp(z)-1)^k
\\ = \frac{n(n-1)}{n^m} \times
m! [z^m]
\sum_{k=2}^n {n-2\choose k-2} (\exp(z)-1)^k
\\ = \frac{n(n-1)}{n^m} \times
m! [z^m] (\exp(z)-1)^2
\sum_{k=0}^{n-2} {n-2\choose k} (\exp(z)-1)^k
\\ = \frac{n(n-1)}{n^m} \times
m! [z^m] (\exp(z)-1)^2 \exp((n-2)z).$$
This is
$$\frac{n(n-1)}{n^m} (n^m - 2(n-1)^m + (n-2)^m)
\\ = n(n-1) \left(1-2\left(1-\frac{1}{n}\right)^m
+ \left(1-\frac{2}{n}\right)^m \right).$$
We get for the variance
$$\mathrm{Var}(N) = \mathrm{E}[N^2] - \mathrm{E}[N]^2
= \mathrm{E}[N(N-1)] + \mathrm{E}[N] - \mathrm{E}[N]^2$$
which yields
$$n^2 \left(1-2\left(1-\frac{1}{n}\right)^m
+ \left(1-\frac{2}{n}\right)^m \right)
\\ - n \left(1-2\left(1-\frac{1}{n}\right)^m
+ \left(1-\frac{2}{n}\right)^m \right)
+ n \left(1-\left(1-\frac{1}{n}\right)^m\right)
\\ - n^2 \left(1-2\left(1-\frac{1}{n}\right)^m
+ \left(1-\frac{1}{n}\right)^{2m} \right)
\\ = n^2 \left(\left(1-\frac{2}{n}\right)^m
- \left(1-\frac{1}{n}\right)^{2m} \right)
+ n \left(\left(1-\frac{1}{n}\right)^m
- \left(1-\frac{2}{n}\right)^m \right).$$
The absolute value of the two differences of terms that are geometric
in $m$ with positive common ratio less than one is bounded by the
maximum of these two, which vanishes in $m$ and hence the variance
goes to zero for $n$ fixed and $m$ going to infinity.
The queried standard deviation is then given by the root of the
variance.
Here is some Maple code to explore these numbers.
ENUM :=
proc(n, m)
option remember;
local ind, data, gf;
gf := 0;
for ind from n^m to 2*n^m - 1 do
data := convert(ind, base, n)[1..m];
gf := gf + v^nops(convert(data, `multiset`));
od;
gf/n^m;
end;
EN_PROB := (n, m) -> subs(v=1, ENUM(n, m));
EN_FM1 := (n, m) -> subs(v=1, diff(ENUM(n, m), v));
EN_FM2 := (n, m) -> subs(v=1, diff(ENUM(n, m), v$2));
FM1 := (n, m) -> n*(1-(1-1/n)^m);
FM2 := (n, m) -> n*(n-1)*(1 - 2*(1-1/n)^m + (1-2/n)^m);
EN_VAR := (n, m) -> - EN_FM1(n, m)^2
+ subs(v=1, diff(v*diff(ENUM(n, m), v), v));
VAR := (n, m) ->
n^2*((1-2/n)^m-(1-1/n)^(2*m)) + n*((1-1/n)^m-(1-2/n)^m);
Addendum. As pointed out by @JMoravitz we may need some
clarification of the reasoning by which the probabilities are
obtained. Note that $k$ distinct outcomes first of all require a
choice of these $k$ values from the $n$ possibilities, for a factor of
${n\choose k}.$ Now we need to match up each of these $k$ values
(think of them as listed sequentially from smallest to largest) with a
subset of the $m$ rolls which may not be empty (this is what prevents
double counting here). This yields the labeled combinatorial class (labeled means EGFs)
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SEQ}_{=k}(\textsc{SET}_{\ge 1}(\mathcal{Z}))$$
or alternatively (compare to Stirling numbers of the second kind)
$$\textsc{SEQ}_{=k}(\textsc{SET}_{=1}(\mathcal{Z})
+ \textsc{SET}_{=2}(\mathcal{Z})
+ \textsc{SET}_{=3}(\mathcal{Z}) + \cdots).$$
We get the generating function
$$\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots\right)^k
= (\exp(z)-1)^k$$
and may then continue as before.
Addendum II. Another combinatorial class we can use here is
$$\textsc{SEQ}_{=n}(\mathcal{E} +
\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z})).$$
This gives the EGF
$$G(z, u) = (1+u(\exp(z)-1))^n
= \sum_{k=0}^n {n\choose k} u^k (\exp(z)-1)^k.$$
On evaluating
$$\left. \frac{\partial}{\partial u} G(z, u) \right|_{u=1}$$
using the second form we get the same sum as before. We can also give
an alternate evaluation
$$\frac{1}{n^m} \times m! [z^m]
\left. \frac{\partial}{\partial u} G(z, u) \right|_{u=1}
= \frac{n}{n^m} \times m! [z^m] \exp((n-1)z) (\exp(z) - 1)
\\ = \frac{n}{n^m} \times m! [z^m] (\exp(nz) - \exp((n-1)z))
= \frac{n}{n^m} (n^m - (n-1)^m)
\\ = n \left(1-\left(1-\frac{1}{n}\right)^m\right).$$
For the next factorial moment we need another derivative which is
$$n(n-1)(1+u(\exp(z)-1))^{n-2} (\exp(z)-1)^2.$$
Set $u=1$ to get
$$n(n-1) \exp((n-2)z)(\exp(2z)-2\exp(z)+1).$$
Extracting coefficients now yields
$$\frac{n(n-1)}{n^m} \left(n^m - 2 (n-1)^m + (n-2)^m\right)$$
and the rest is as in the first version. Note that we get a faster
enumeration routine if we admit the classification from the
introduction. The result is midway between enumeration and the closed
form and is shown below.
with(combinat);
ENUM2 :=
proc(n, m)
option remember;
local gf, part, psize, mset;
gf := 0;
part := firstpart(m);
while type(part, `list`) do
psize := nops(part);
mset := convert(part, `multiset`);
gf := gf + binomial(n, psize)
*m!/mul(p!, p in part)
*psize!/mul(p[2]!, p in mset)*v^psize;
part := nextpart(part);
od;
gf/n^m;
end;
I think you have a good hang of the concept. However, things can always be written better.
For example, the sample space for three independent dice, rather than the suggestive $\{(1,1,1),...,(6,6,6)\}$(which is correct, so credit for that) can be written succinctly as $\Xi \times \Xi \times \Xi$, where $\Xi = \{1,2,3,4,5,6\}$. This manages to express every element in the sample space crisply, since we know what elements of cartesian products look like.
The sigma field is a $\sigma$-algebra of subsets of $\Omega$. That is, it is a set of subsets of $\Omega$, which is closed under infinite union and complement.
Ideally, the sigma field corresponding to a probability space, is the set of events which can be "measured" relative to the experiment being performed i.e. it is possible to assign a probability to this event, with respect to the experiment being performed. What this specifically means, is that based on your experiment, your $\sigma$-field can possibly be a wise choice.
In our case, we have something nice : every subset of $\Omega$ can be "measured" since $\Omega$ is a finite set (it has $6^3 = 216$ elements) hence the obvious candidate, namely the cardinality of a set can serve as its measure(note : this choice is not unique! You can come up with many such $\mathcal F$).
Therefore, $\mathcal F = \mathcal P(\Omega)$, where $\mathcal P(S)$ for a given set $S$ is the power set of $S$, or the set of all subsets of $S$. This is logical since this contains every subset of $\Omega$, and is obviously a $\sigma$-algebra.
Now, $P(A)$, for $A \subset \mathcal F$(equivalently, $A$ any subset of $\Omega$) is, for the reasons I mentioned above, simply the ratio between its cardinality, and the cardinality of $\Omega$. Therefore, $P(A) = \frac{|A|}{216}$. That is exactly what you wrote, except well, division by sets isn't quite defined.
So there you have it, an answer, along with what you've done right and wrong.
NOTE : $\mathcal P$ for the power set, and $P$ for the probability of a set is my notation here. I still think the letter $P$ is used for both, but it's causing confusion here, hence the change.
Best Answer
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.