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.
The answer to your question is positive and follows from Theorem 6.4.32 in Qing Liu's book Algebraic geometry and arithmetic curves.
Note that Liu uses Corollary 6.4.13 in the statement of his Theorem. Moreover, the base scheme is a locally Noetherian scheme, e.g., the spectrum of a field.
Best Answer
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 $S_2$, $$\mathcal Hom_X(\omega_X,\omega_X)\simeq ({\omega_X\otimes \omega_X^*})^{**}\simeq \mathcal O_X.$$