Do the categories of Sets and Smooth manifolds with smooth functions have zero morphisms. And how are compositions with the empty set defined? When reading wiki zero morphisms it seems that the category of sets does not have zero morphisms. Also, I could not find how the composition with the empty map works in the Set. Can someone explain that?
Edit: there is a confusion about what 'does have zero morphisms' mean. Here I actually mean: 'category with zero morphisms'.
 A: Thomas Andrews has answered your main question about zero morphisms, so I'll just answer the question "how are compositions with the empty set defined?"
In the category of sets, a function $X\to Y$ is a set $f\subseteq X\times Y$ such that for all $x\in X$, there is a unique $y\in Y$ such that $(x,y)\in f$. From this definition, we see that $\emptyset \subseteq X\times Y$ is a function $X\to Y$ if and only if $X = \emptyset$. Indeed, if $X\neq \emptyset$, then picking some $x\in X$, there is no $y\in Y$ such that $(x,y)\in \emptyset$, so $\emptyset$ is not a function $X\to Y$. On the other hand, if $X = \emptyset$, then $\emptyset$ satisfies the definition of a function $X\to Y$ vacuously. 
Now how do we compose with the empty function? Well, let's view $\emptyset$ as a function $\emptyset \to Y$, and suppose we have a function $g\colon Y\to Z$. Composing, we should get a function $(g\circ \emptyset)\colon \emptyset \to Z$. You should expect to get $(g\circ \emptyset) = \emptyset$, since the empty function is the only function with domain $\emptyset$. And indeed, we have $$(g\circ \emptyset) = \{(x,z)\mid \exists y\, (x,y)\in \emptyset\text{ and }(y,z)\in g\} = \emptyset.$$
On the other hand, let's view $\emptyset$ as a function $\emptyset \to Y$, and suppose we have a function $g\colon Z\to \emptyset$. Composing, we should get a function $(\emptyset\circ g)\colon Z\to Y$. Again, we have $$(\emptyset\circ g) = \{(z,y)\mid \exists x\, (z,x)\in g\text{ and }(x,y)\in \emptyset\} = \emptyset.$$
In fact, in this case we must have $Z = \emptyset$ and $g = \emptyset$, since there are no functions from non-empty sets to the empty set. 
A: Some categories have initial and final object isomorphic (call it $0/1.$) Then for any objects $X,Y,$ you have a unique $o_{XY}:X\to 0/1\to Y$ which you can show is a zero-morphism and it satisfies the requirement to be a category "with zero morphisms."
This class contains the typical examples of categories with zero morphisms:


*

*Groups, R-modules, Vector spaces over a fixed field

*Pointed sets, pointed topological spaces


In Sets, if $X\neq\emptyset$ and $Y=\emptyset,$ there is no map $X\to Y,$ so there are not enough maps for Sets to be a "category with zero morphisms."
I think you can show that any map in Smooth manifolds fails to qualify as a zero-morphism. Basically, if $h:M_1\to M_2$ with $M_1$ non-empty, then we can choose some pair $f,g:M_2\to M_3$ so that $f,g$ are not equal on the image of $h$ in $M_2.$ 
For example, if the category includes disconnected manifolds, we can take $M_3=M_2\sqcup M_2$ and $f,g$ the two natural maps $M_2\to M_2\sqcup M_2.$
It might be a little harder if the category consisted only of connected manifolds.
