(a) Let us define the functions \\( f : t \rightarrow t' \\) and \\( g : t' \rightarrow t \\)
Their composition gives us \\( g \circ f : t \rightarrow t \\) and \\( f \circ g : t' \rightarrow t' \\)
Since there is only one morphism from any \\( a \\) to \\( t \\) or \\( t' \\) then these two compositions must necessarily be the identity function.


(b) Since \\( p, p' \\) are both products of \\( a, b \\) then there is a unique morphism from \\( c \\) to them ( \\( h, h' \\) ). If t we have two morphisms \\( \alpha : p \rightarrow p' \\) and \\( \beta : p' \rightarrow p \\) . Since \\( h, h' \\) are unique, we have:

$$
\alpha \circ h : c \rightarrow p' = h'
$$
$$
\beta \circ h' : c \rightarrow p = h
$$
$$
\alpha \circ \beta \circ h' = h'
$$
$$
\beta \circ \alpha \circ h = h
$$

And our only choice is for \\( \alpha, \beta \\) be the identity functions.

(c) both proofs used the uniqueness of morphisms and the commuting diagrams. I think we could follow the same idea for the initial objects, since there is only one morphism from \\( 0 \rightarrow \ \\).