The dualizing sheaf looks like the sheaf of 1-forms, with logarithmic singularities at nodes (of the form f(z)dz/z, f regular at 0) such that the residues on each component of a node add to zero. The answer to the second part of number 1 is "yes" but you can also just calculate using the explicit description.
The answer is no.
$\mathcal Hom (\mathcal O_X,\omega)\simeq \omega$, but $\omega\otimes \mathcal Hom (\omega, \mathcal O_X)$ is not necessarily reflexive, let alone locally free. However, it is true that
if $X$ is $G_1$, that is, Gorenstein in codimension $1$, then
$$
(\omega\otimes \mathcal Hom (\omega, \mathcal O_X))^{**}\simeq \mathcal O_X.
$$
EDIT: (due to popular demand, here is a better formed statement for the CM is not even needed part of the original)
This is actually true in a little more general setting:
For simplicity assume that $X$ is $G_1$, $S_2$, equidimensional of dimension $d$ and admits a dualizing complex denoted by $\omega_X^\bullet$. (If, say, by variety you mean a quasi-projective (reduced) scheme of finite type over a field, then the last assumption is automatic. If in addition you also mean irreducible, then so is the equidimensionality. CM obviously implies $S_2$.)
Let
$$\omega_X := h^{-d}(\omega_X^\bullet)$$
and
$$\omega_X^*:=\mathcal Hom_X (\omega_X, \mathcal O_X).$$
Then
$${(\omega_X\otimes \omega_X^*)}^{**}\simeq \mathcal O_X.$$
This follows by the fact that $X$ is $S_2$, both sides are reflexive and they agree in codimension $1$ due to the $G_1$ assumption.
EDIT2: (inspired by Karl's answer):
This actually also implies that
$$\mathcal Hom_X(\omega_X,\omega_X)\simeq \mathcal O_X$$
(under the same conditions) since on the open set where $\omega_X$ is a line bundle,
$$\mathcal Hom_X(\omega_X,\omega_X)\simeq {\omega_X\otimes \omega_X^*}$$
and then since they are both reflexive and $X$ is $S_2$,
$$\mathcal Hom_X(\omega_X,\omega_X)\simeq ({\omega_X\otimes \omega_X^*})^{**}\simeq \mathcal O_X.$$
Best Answer
With regards part 2.
Let's assume that you have two components $X_1$ and $X_2$ (or even unions of components) such that $X_1 \cup X_2 = X$=. Let $I_1$ and $I_2$ denote the ideal sheaves of $X_1$ and $X_2$ in $X$.
Set $Z$ to be the scheme $X_1 \cap X_2$, in other words, the ideal sheaf of $Z$ is $I_1 + I_2$.
It is easy to see you have a short exact sequence $$0 \to I_1 \cap I_2 \to I_1 \oplus I_2 \to (I_1 + I_2) \to 0$$ where the third map sends $(a,b)$ to $a-b$.
The nine-lemma should imply that you have a short exact sequence
$$0 \to O_X \to O_{X_1} \oplus O_{X_2} \to O_Z \to 0$$
If you Hom this sequence into the dualizing complex of $X$, you get a triangle $$\omega_Z^. \to \omega_{X_1}^. \oplus \omega_{X_2}^. \to \omega_{X}^. \to \omega_Z^.[1]$$
You can then take cohomology and, depending on how things intersect (and what you understand about the intersection), possibly answer your question.
If $X_1$ and $X_2$ are hypersurfaces with no common components (which should imply everything in sight is Cohen-Macualay) then these dualizing complexes are all just sheaves (with various shifts), and you just get a short exact sequence $$0 \to \omega_{X_1} \oplus \omega_{X_2} \to \omega_{X} \to \omega_{Z} \to 0$$
Technically speaking, I should also probably push all these sheaves forward onto $X$ via inclusion maps.