2) **0** is the category with no objects or morphisms (i.e., both sets are \$$\emptyset\$$.

3) Claim: There are \$$\frac{n(n+1)}{2}\$$ morphisms in **n** for all \$$n\in\mathbb{N}\$$.
Proof: For every two objects, say \$$p>q\$$, in **n** there is exactly one one morphism, that obtained from the compositions \$$f_{p-1}\circ f_{p-2}\circ\cdots\circ f_{q+1}\circ f_q\$$. By definition of **n**, these constitute all of the non-identity morphisms in **n**. So, there are \$$\binom{n}{2}=\frac{n(n-1)}{2}\$$ non-identity morphisms and \$$n\$$ identity morphisms, for a total of \$$\frac{n(n+1)}{2}\$$ morphisms.