Monomorphisms are embeddings functors (i.e. functors whose both the object-function and the arrow-function are injective).
It easily seen that embeddings are monomorphisms because if $F \colon \mathbf C \to \mathbf D$ is an embedding then for every pair of fucntors $G,H \colon \mathbf X \to \mathbf C$ such that $F \circ G=F \circ H$ then
for every $x \in \mathbf X$ we should have $F \circ G(x)=F \circ H(x)$ and since $F$-object part is injective it would follow that $G(x)=H(x)$;
in exactly the same way for every $f \colon x \to x'$ in $\mathbf X$ we have that $F \circ G(f)=F\circ H(f)$ and for the injectivity of $F$-arrow part we also have that $G(f)=H(f)$;
so $G=H$.
Now let's suppose $F \colon \mathbf C \to \mathbf D$ is monomorphism.
For every object $c \in \mathbf C$ there's a unique functor $\bar x \colon 1 \to \mathbf C$ (where $1$ is the category with one object ($0$) and one arrow) such that $\bar x(0)=x$.
In similar way for every morphism $f \in \mathbf C(c,c')$ there's a unique functor $\bar f \colon 2 \to \mathbf C$ (where $2$ is the category with two objects, $0$ and $1$, and only one non identity morphism $0 \to 1$) such that $\bar f(0)=c$, $\bar f(1)=c'$ and $\bar f(0 \to 1)=f$.
Then it follows that for every pair of objects $c,c' \in \mathbf C$ if we have $F \circ \bar c(0)=F(c)=F(c')=F \circ \bar c'(0)$ then $F \circ c=F \circ c'$ and so by monomorphism property $\bar c=\bar c'$. From this follows that
$$c=\bar c(0)=\bar c'(0)=c'$$
which gives the injectivity of the object part of $F$.
If $f,g \colon c \to c'$ are morphisms of $\mathbf C$ such that $F(f)=F(g)$ it can be proven similarly (using the functor $\bar f$ and $\bar g$) that $f=g$.
From this it follows that also the arrow part of $F$ is injective.
Edit I see that I didn't address the part about epimorphisms in $\mathbf{Cat}$. Let's try to make ammend.
For start it can be shown that every functor that is not sujective on objects cannot be an epimorphism. Indeed if $F \colon \mathbf C \to \mathbf D$ is not surjective we can consider the full subcategory $\mathbf D' \hookrightarrow \mathbf D$ that is spanned by the object in $\text{Im} F$. The consider the pushout of the diagram
where the morphism from $\mathbf D'$ are the embeddings.
Since $F$ factors through $i \colon\mathbf D' \hookrightarrow \mathbf D$ (since it's image is contained in the full subcategory $\mathbf D'$) we have that $G \circ F = H \circ F$.
More in detail there's a functor $\mathcal F \colon \mathbf C \to \mathbf D'$ such that $F=i \circ \mathcal F$ and so
$$G \circ F = G \circ i \circ \mathcal F$$
and
$$H \circ F = H \circ i \circ \mathcal F\, .$$
Since $G \circ i= H \circ i$, because of the pushout property, we have that $G \circ F=H \circ F$, nonetheless $G \ne H$.
This basically prove that epimorphism should at least be sujective on objects.
Clearly this condition is not sufficient, it's easy to find counter examples.
What can be proven is that $F \colon \mathbf C \to \mathbf D$ is sujective on objects and the graph $\text{Im}F$ contains a set of morphism which are generators for the category $\mathbf D$ then this functor is an epimorphism.
By generators for the category $\mathbf D$ I mean a family of morphisms such that every other morphism in $\mathbf D$ is a composite of these morphisms.
The proof of this is no different from the analog proofs for other categories.
Nontheless this is just a sufficient condition as the following example show.
Consider the category $2$ defined as above and $\bar 2$ obtained by $2$ just adding an inverse to the morphism $0 \to 1$.
The embedding $j \colon 2 \to \bar 2$ is an epimorphism since for every $G,H \colon \bar 2 \to X$ such that $G(0\to 1)=G(0 \to 1)$ have to send $1 \to 0$ in *the inverse of $G(0 \to 1)=H(0 \to 1)$.
Of course from this example it becomes clear that one can perfect the previoius said condition by proving that $F \colon \mathbf C \to \mathbf D$ is an epimorphism iff it's surjective on object and every morphism in $\mathbf D$ is a composite of morphism in $\text{Im} F$ and their inverse (if they exist).
Nonetheless I'm not aware if this is a full characterization of epimorphisms in $\mathbf{Cat}$.
Hope this helps.
Remember that you don't have to define Hom functors from scratch; you should already know how they work. Given $\langle f,g,h\rangle:\langle A,B,C\rangle\to\langle A',B',C'\rangle$, $F(f,g,h)(a:A\times B\to C)$ is just $a\mapsto h\circ a\circ (f\times g)$. (Notice that you have the variance backwards above--the first argument of a Hom functor is contravariant, the second covariant.) $G$ requires a little more thought, but it's easy to sort out with a little time.
$H$ is, I think, where you're getting confused. All you need to show is that for every $A,B,C$ there's a function between sets $H_{A,B,C}:F(A,B,C)\to G(A,B,C)$ that commutes with the $F(f,g,h)$ and the $G(f,g,h)$. That's it. You don't need any extra functors. These functions are given by the currying operation described above.
Best Answer
No, elements of $Cat$ are small categories - $Cat\times Cat$ is the domain, and $Cat$ the codomain, of $\times$. Analogously, "a function on natural numbers" would be $f:\mathbb{N}\rightarrow\mathbb{N}$ but that wouldn't imply that $\mathbb{N}$ is a natural number.
Indeed, $Cat$ itself is not a small category, since there are a proper class of small categories (easy exercise: there is a category of cardinality $\kappa$ for every cardinal $\kappa$).
(Of course, $Cat$ is locally small - given any two small categories there are only set-many functors between them - but it's not small.)