For 1. If you are willing to accept facts about the bar construction, consider $A = \Sigma^{-1} V\oplus k$ and define a commutative algebra structure on $A$ by insisting that the product on $\Sigma^{-1} V$ is zero. This makes $A$ into a commutative algebra. Now, we have $BA$ the bar construction on $A$, and $BA$ is the same as your $TV$. A map $C \to BA$ of graded coalgebras is completely determined by the projection $C\to BA \to A$ of degree $-1$. Given a map $f: C\to A$ of degree $-1$, it extends to a coalgebra map $C\to BA$ provided that it satisfies $0 = m(f\otimes f)\Delta$ where $m$ is the product on $A$, and $\Delta$ is the coproduct on $C$. Now, check that if $A$ is a commutative algebra, then the map $BA\otimes BA \to A$ given by $[a]\otimes 1 \mapsto a$, and $1\otimes [a] \mapsto a$, and zero on all other tensor factors, satisfies the condition. Then, check that the induced map $BA\otimes BA \to BA$ is the shuffle product. Similarly, check that it is associative by projecting to $A$.
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