First a comment: You have to be a little bit careful with the translation from the concrete case, since the categorical definition of subobject (in terms of monic arrows) may not coincide with "subset which is also an object", even in a concrete category. Also, even if subobjects have their "usual" meaning in a concrete category, it seems Theorem 1 may not hold. See the counterexample at the end of the answer, which takes place in a concrete category.
Here's a simpler version of your question:
Let $X$ be an object, $m\colon X'\hookrightarrow X$ a subobject, and $f\colon Z\to X$ an arbitrary arrow. We say $f$ factors through $X'$ if there is an arrow $f'\colon Z\to X'$ such that $m\circ f' = f$. Note that since $m$ is monic, $f'$ is unique if it exists.
Question 1': Let $C$ be a category with Properties 1 and 2 (products and intersections of subobjects). Let $X$ be an object, $(X_i)_{i\in I}$ a family of subobjects of $X$, and $X_\cap$ their intersection. Let $f\colon Z\to X$ be an arbitrary arrow which factors through each subobject $X_i$. Does $f$ factor through $X_\cap$?
I claim that Question 1' is equivalent to your Question 1.
Suppose Question 1' has a positive answer. Then for any arrow $f\colon Z\to X$, we can define $\text{im}_f(Z)$ to be the intersection of the family of all subobjects $X'\hookrightarrow X$ such that $f$ factors through $X'$. Then $f$ factors through $\text{im}_f(Z)$, so we have $Z\to \text{im}_f(Z)\hookrightarrow X$.
Now given $m\colon Z\hookrightarrow X\times Y$ as in Question 1, define $f = \pi_X\circ m\colon Z\to X$ and $g = \pi_Y\circ m\colon Z\to Y$. Then it's straightforward to show that $\text{im}_f(Z)\times \text{im}_g(Z)\hookrightarrow X\times Y$ is a $Z$-box, and moreover it is below every other $Z$-box, so it is the intersection of all $Z$-boxes.
Conversely, suppose Question 1 has a positive answer. Let $X$ be an object, $(X_i)_{i\in I}$ a family of subobjects of $X$, $X_\cap$ their intersection, and $f\colon Z\to X$ an arrow which factors through each $X_i$. Note that $f\times \text{id}_Z\colon Z\to X\times Z$ is a monic arrow, and for any subobject $m\colon X'\hookrightarrow X$ such that $f$ factors through $X'$, we have that $m\times \text{id}_Z\colon X'\times Z\hookrightarrow X\times Z$ is a $Z$-box. By Question 1, the intersection of all $Z$-boxes is a $Z$-box, which has the form $g\times \text{id}_Z\colon X'\times Z\hookrightarrow X\times Z$. Then $f$ factors through $X'$, and $g\colon X'\hookrightarrow X$ is below $X_i$ for all $i\in I$, hence $X'$ is below $X_\cap$, and thus $f$ factors through $X_\cap$.
Now in "normal" situations, the intersection $X_\cap$ of a family of subobjects will actually be the limit of the diagram consisting of all the monic arrows $m_i\colon X_i\hookrightarrow X$ (this is a fiber product / pullback of a family of monic arrows). In this case, Question 1' has a positive answer, by the universal property of the limit. Conversely, if this diagram has a limit, then the limit object will be a subobject of $X$, and hence will be the intersection of the family.
So a natural answer to Question 3 is: Claim 2 is true in any category which has fiber products of families of monic arrows.
On the other hand, to answer Question 2: We can refute Question 1' by exhibiting a family of subobjects which has an intersection but fails to have a fiber product. Consider the category of all groups which do not have size $2$. This category has products (since no product of groups of size $\neq 2$ has size $2$) and intersections of subobjects: If a family of subgroups of a group has a set-theoretic intersection of size $2$, then their category-theoretic intersection is the trivial group.
Now let $X = C_2\times \mathbb{Z}\times \mathbb{Z}$, let $X_1 = C_2\times \mathbb{Z}\times \{0\}$, and let $X_2 = C_2\times \{0\}\times \mathbb{Z}$. Let $Z = \mathbb{Z}$, and let $f\colon Z\to X$ be the map $f(n) = (\overline{n},0,0)$, where $\overline{n}\in C_2$ is the reduction of $n$ mod $2$. Then $f$ factors through $X_1$ and $X_2$, but it fails to factor through the category-theoretic intersection of $X_1$ and $X_2$, which is the trivial group.
This is the usual argument how to recover a universal property from a representable presheaf. Plugging in $\Omega$ into the isomorphism of functors yields a universal subobject $t$ that corresponds to the identity on $\Omega$, and then the desired universality follows immediately since every map into $f\colon E\to\Omega$ is the image of the identity on $\Omega$ along $f^\ast$ and since $f^\ast$ acts on subobjects by means of pullback along $f$.
Edit: To see that the domain $X$ of $t$ is the terminal object, note that any morphism $f\colon E\to X$ determines a map $tf$ and therefore a subobject of $E$. By construction, $E$ must be a retract of this subobject, which is only possible if the subobject is the maximal one (i.e. the identity on $E$). As the identity on $E$ defines a subobject of $E$ and therefore gives rise to such a map $f$, this shows that there is a unique such map $f$, hence $X$ is terminal.
Best Answer
Ok, my previous answer wasn't correct. Now, everything works.
First of all, I thank Sergio Buschi, who provided me a counter-example to the OP. Since he gave me his permission, I'll write the details. Let $\mathcal C$ be the subcategory of $\mathbf{Set}$ whose objects are all the sets. As morphisms, we include the two natural morphisms $u,v \colon \{0\} \to \{0,1\}$; in addition, for every other set $S$ we choose exactly one morphism $t_S \colon \{0\} \to S$ and we add exactly one morphism $S \to \{0,1\}$ sending every element to $0$. Then the morphisms $S \to \{0,1\}$ are monomorphisms and $\{0\}$ is a strong generator (easy check). Hence $\mathcal C$ is not well-powered.
As for my problem, I realized two things:
I must again thank Sergio Buschi for the terse statement above. I gave a proof by myself, but in stating it I was assuming too much hypothesis.