We only have to use the definitons of the universal properties we want to prove.
(1) (Coproduct part)
I suppose that you are assuming that the $A_i$ are disjoint, if not WLOG we can find a family of disjoint objects wich are isomorphic to each of the previous ones. Let $B$ be an object in the cathegory (a set), with a family of morphisms (maps) $\{f_i:A_i\rightarrow B\}$, we have to show now that there is a unique $f:C\rightarrow B$ such that for each $i$, $f\circ c_i=f_i$, i.e. making the corresponding diagram conmutative.
For existence, we define a function $f:C\rightarrow B$ as follows
$$f(a)=f_i(a)$$
whenever $a\in A_i$. It is immediate to see that $f$ is well defined since the $A_i$ are disjoint, and that $f\circ c_i(a)=f_i(a)$ for all i and $a\in A_i$.
For uniqueness, suppose that there is other $g:C\rightarrow B$ with the same property; then since $g$ and $f$ have the same codomain and domain we only have to shown that $g(x)=f(x)$ for all $x\in C$. So let $x\in C$, then $x\in A_i$ only for one $i$ obtainig that
$$f(x)=f_i(x)=g\circ c_i(x)=g(c_i(x))=g(x)$$
as desired.
Hence, the disjoint union of the $A_i$ is the desired coproduct in the cathegory $\mathcal{C}_1$.
(Product part)
For the product, take an object in the desired cathegory (a set) $B$ with a family of mophisms $\{f_i:B\rightarrow A_i\}$. Then, we have to show that there is a unique $f:B\rightarrow P$ such that for each $i$, $p_i\circ f=f_i$, that as the previous case means making the corresponding diagram conmutative.
For existence of such an $f:B\rightarrow P$, define
$$f(b)=(f_1(b),...,f_n(b))$$
for each $b\in B$. Then, with any problem we see well-defineness and it can be seen that for each $b\in B$ and $i$ we have
$$p_i\circ f (b)=p_i(f(b))=p_i((f_1(b),...,f_n(b)))=f_i(b)$$
as desired.
For uniquenes, take $g:B\rightarrow P$ with the same property. Now, given a $b\in B$ for each $i$ we have that
$$p_i\circ f (b)=f_i(b)=p_i\circ g(b)$$
So, the tuples $f(b)$ and $g(b)$ in the cartesian product have each of it componets equal and thus they must be equal. From this we have that for each $b\in B$, $f(b)=g(b)$, and since $f$ and $g$ share the codomain and domain they must be the same function.
Hence, the cartesian product of the $A_i$ is the desired product in the cathegory $\mathcal{C}_1$.
For (2) and (3) the previous proofs works perfectly, we have only to add the technical detail of shaw that $f:B\rightarrow P$ is in that case the corresponding linear map of homomorphism. Then, the rest of set theoric details continue the same.
EDIT: (I will add now the rest of the proofs.)
(2) First of all, we will notice that for the finite case of the family as noted by Arturo Magidin the direct sum of the $A_i$ and their cartesian product is the same -which in the infinite case is not true, thnaks that when the cartesian product the finiteness condition of tuples is useless-.
One, we have make this observation we can center in the cartesian product with vector space structure for our proof for similarity with other cases.
(Coproduct part)
First, observe that the $c_i$ are linear maps. Once we have done that observation, we consider our object (in this case a vector space) $B$ with a family of morphisms (linear maps) $\{f_i:A_i\rightarrow B\}$. As before, we have to show that there is a unique morphism $f:\prod A_i\rightarrow B$ such that $f\circ c_i=f_i$.
For existence, consider $f: \prod A_i\rightarrow B$ given by
$$f((a_1,...,a_n))=f_1(a_1)+...+f_n(a_n)$$
where $a_i\in A_i$. (See why this trick fails in general in non-abelian groups, and why is important conmutativity for it.) Then, we comprobe inmediately that for any $i$ and each $a_i\in A_i$ we have
$$f\circ c_i(a_i)=f(c_i(a_i))=f((0,...,a_i,...,0))=f_1(0)+...+f_i(a_i)+...+f_n(0)=f_i(a_i)$$
as desired.
For uniqueness, consider a $g: \prod A_i\rightarrow B$ with he same property. Then, we hva for each $(a_1,...,a_n)\in\prod A_i$ that (by linearity)
$g((a_1,...,a_n))=g(\sum (0,...,a_i,...,0))=\sum g((0,...,a_i,...,0))=\sum g(c_i(a_1,...,a_n))=\sum f_i(a_i)$
So, since $f((a_1,...,a_n))=\sum f_i(a_i)$ we obtain $f((a_1,...,a_n))=g((a_1,...,a_n))$. From where by cinciden of codomain and domain, $f=g$ as wanted.
(Product part)
We only have to apply the result in the cathegory of sets, by showing that in the standard case of an object B and a family of morphisms $\{f_i:B\rightarrow A_i\}$ the map
\begin{align}
f:B&\rightarrow \prod A_i\newline
b&\mapsto (f_1(b),...,f_n(b))
\end{align}
is a linear transformation. Which is a basic result of linear algebra. Then we apply the previous set theoric conclusions thanks to the cathegory being concrete.
(3) The same a for vector spaces. The basic fact is that in that the candidate for product is the same that the one in the cathegory of sets, but adding some extra information to the set and that the candidate set theoric map
\begin{align}
f:B&\rightarrow \prod A_i\newline
b&\mapsto (f_1(b),...,f_n(b))
\end{align}
is a morphism -homomorphism- in this cathegory.
Best Answer
Let's prove that the coproduct is unique in a category. Assume that there are two, let's say $Q$ and $Q'$ for the collection $\{A_a\}$. Let $\pi_a:A_a\to Q$ and $\pi':A_a\to Q'$ be the morphisms you talked about. Then by the universal property you mentioned, we have that there are morphisms $\phi:Q'\to Q$ and $\psi:Q\to Q'$ such that $\pi_a=\phi\circ\pi_a'$ and $\pi_a'=\psi\circ\pi_a$. Now these morphisms $\phi$ and $\psi$ are unique. If we now take $\phi\circ\psi:Q\to Q$, this also satisfies $\phi\circ\psi\circ\pi_a=\phi\circ\pi_a'=\pi_a$, the same as the identity. Since the map of the universal property is unique, we have that $\phi\circ\psi=\mbox{id}$. The same argument shows that $\psi\circ\phi=\mbox{id}$.
So the coproduct is unique. Therefore, if you can show that the direct sum of abelian groups satisfies what you are looking for (it seems you already did this?), then it must be "the" coproduct (up to isomorphism).