Yes. This follows from the strict topology version of the Kaplansky Density Theorem (the proof is much the same as the usual proof of Kaplansky Density). See for example Proposition 1.4 in Lance's book about Hilbert C*-modules. In multiplier algebra language it says:

Theorem: Let $A$ be a C-algebra and let $B$ be a strictly dense C-subalgebra of $M(A)$. Then the unit ball of $B$ is strictly dense in the unit ball of $M(A)$.

It is easy to see that $B$ need not be closed, as in your example.

Yes. This follows from the strict topology version of the Kaplansky Density Theorem (the proof is much the same as the usual proof of Kaplansky Density). See for example Proposition 1.4 in Lance's book about Hilbert $C^*$-modules. In multiplier algebra language it says:

Theorem: Let $A$ be a $C^*$-algebra and let $B$ be a strictly dense $C^*$-subalgebra of $M(A)$. Then the unit ball of $B$ is strictly dense in the unit ball of $M(A)$.

It is easy to see that $B$ need not be closed, as in your example.