Let $(\Omega,\mathcal{F},P)$ be a probability space. Suppose $X$ is an integrable random variable and let $\mathcal{G}$ be a sub-sigma-field of $\mathcal{F}$. The conditional expection $E[X\mid\mathcal{G}]$ is the unique random variable that satisfies:
1) $E[X\mid\mathcal{G}]$ is $\mathcal{G}$-measurable.
2) $\int_A E[X\mid\mathcal{G}]\,\mathrm dP = \int_A X\,\mathrm dP$ for all $A\in\mathcal{G}$.
I'm assuming you want to find expressions for $E[X\mid\mathcal{F}]$ and $E[X\mid\{\emptyset,\Omega\}]$. For the first conditional expectation, try showing that $X$ satisfies 1) and 2), and for the last conditional expectation, try with $E[X]$.
One some times sees $E[X\mid\mathcal{G}]$ as our best guess of $X$ given the information contained in $\mathcal{G}$. Try holding this up with the expectations when $\mathcal{G}=\mathcal{F}$ and $\mathcal{G}=\{\emptyset,\Omega\}$.
Let $\cal H$ be the $\sigma$-algebra generated by ${\Bbb E} (X\mid\mathcal G)$.
As you say, since ${\Bbb E} (X\mid\mathcal G)$ is $\mathcal G$-measurable, $\cal H$ is contained in $\mathcal G$. It doesn't have to be contained in $\sigma(X)$. For a simple counterexample, let the probability space be $\Omega=\{1,2,3,4\}$ with all subsets measurable and the uniform probability measure, and let $${\mathcal G}=\{\emptyset,\{1,2\},\{3\},\{4\},\{3,4\},\{1,2,3\},\{1,2,4\},\Omega\},$$ $$X(1)=1, X(2)=X(3)=X(4)=0.$$ Then
$$
{\Bbb E} (X\mid\mathcal G)(1)={\Bbb E} (X\mid\mathcal G)(2)=\frac12,
\ \ \ \
{\Bbb E} (X\mid\mathcal G)(3)={\Bbb E} (X\mid\mathcal G)(4)=0$$
so ${\cal H}=\{\emptyset,\{1,2\},\{3,4\},\Omega\}$ is not contained in $\sigma(X)=\{\emptyset,\{1\},\{2,3,4\},\Omega\}$.
Since $\cal H$ does not have to be contained in $\sigma(X)$, it need not equal ${\cal G}\cap\sigma(X)$. The example above shows that it also does not have to equal $\cal G$.
Addendum: As Byron Schmuland says below, if $X$ is an ${\cal F}/{\cal B}({\Bbb R})$-measurable real-valued function, then $\cal H$ is countably generated as it's the pullback of the countably generated $\sigma$-algebra ${\cal B}(\Bbb R)$ under ${\Bbb E} (X\mid\mathcal G)^ {-1}$. Also, any countably generated $\sigma$-algebra which is contained in $\cal G$ can occur as a value of $\cal H$ for some real-valued $X$. (${\Bbb E} (X\mid\mathcal G)$ is only defined up to equality on a set of measure $1$, so to be precise you should say that any countably generated $\sigma$-algebra which is contained in $\cal G$ can occur as a value of $\sigma({\Bbb E}(X\mid\mathcal G))$ for some real-valued $X$ and some version of ${\Bbb E}(X\mid \mathcal G)$.)
Best Answer
The conditional expectation $E(X|\mathcal G)$ is a random variable determined (up to sets of probability $0$) by two properties: (i) $E(X|\mathcal G)$ is $\mathcal G$-measurable, and (ii) for each $\mathcal G$-measurable set $B$, the integral of $E(X|\mathcal G)$ over $B$ is equal to the integral of $X$ over $B$. If $X$ is itself $\mathcal G$-measurable, then it has these two properties, and so it must be $E(X|\mathcal G)$.