A polynomial $f(x)$ with coefficients in any field $K$ and of degree $\leq3$ is irreducible over $K$ if and only if it has no roots in $K$. The given polynomials have no roots in $K=\Bbb Z_{11}$ and are therefore irreducible over $\Bbb Z_{11}$.
A standard fact about the ring of polynomals $K[X]$ is that its ideals are always principal, i.e. generated by one element. A standard consequence is that the ideals generated by irreducible polynomials are maximal. In addition, a quotient of a ring by an ideal is a field if and only if that ideal is maximal.
This explains why the given quotients are fields.
If $f(X)$ is irreducible of degree $d$, a full set of representants of the quotient $K[X]/(f(X))$ is given by the set of polynomials of degree $\leq d-1$ (this is an easy exercise). It follows that if $K$ is a finite field with $q$ elements the said quotient has $q^d$ elements. This explains why the given quotients have $121=11^2$ elements.
Finally, the general theory of finite fields tells us that for any prime power $q=p^f$ there exists a field with $q$ elements, which is unique up to isomorphism. Basically, the uniqueness follows from the fact that a finite field with $q=p^f$ elements is made up of the roots of the polynomial $X^q-X$ in some chosen algebraic closure of the basic field $\Bbb F_p=\Bbb Z/\Bbb Zp$.
Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to
$$
\frac{f(m)}{f(n)}\in F
$$
where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then
$$
n=\underbrace{1+1+\dots+1}_{\text{$n$ times}}
$$
and therefore $0\prec n$. Conversely, if $n<0$, then
$$
n=-(\,\underbrace{\,1+1\dots+1}_{\text{$-n$ times}}\,)
$$
and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields,
$$
0\prec n\cdot\frac{m}{n}=m
$$
and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).
Best Answer
Note that the zeros of $x^{p^n}-x$ are exactly the elements of a finite field with $p^n$ elements. Since splitting fields are uniquely determined up to isomorphism, we can write $GF(p^n)$.
First, if $m$ divides $n$, then $p^m-1$ divides $p^n-1$ and moreover $x^{p^n-1}-1$ divides $x^{p^n-1}-1$. Thus $x^{p^m}-x$ divides $x^{p^n}-x$. Hence, $GF(p^m)$ is a subfield of $GF(p^n)$.
Conversely, if $GF(p^m)$ is a subfield of $GF(p^n)$, then $GF(p^n)$ is a vector space over $GF(p^m)$. Thus $p^n = (p^m)^k$ for some $k\geq 1$. Hence, $m$ is a divisor of $n$.
Added comment: Suppose $m$ divides $n$ with $n=km$. Then $\frac{p^n-1}{p^m-1}= \frac{(p^m)^k-1}{p^m-1} = (p^m)^{k-1} + (p^m)^{k-2} +\ldots+p^m+1$.