(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 \ \$$.