No. Take some element $f$ of $K[x]$ which is not in the ideal $(x)$ and is not invertible. Then $$K[[x]] \otimes_{K[x]} K[x]/(f) \cong K[[x]]/(f) = 0$$ where the latter follows from the fact that since $f \notin (x)$, it becomes invertible in $K[[x]]$.

In particular, $K[[x]]$ is not faithfully flat over $K[x]$, so it cannot be free over it.

No. Take some element $f$ of $K[x]$ which is not in the ideal $(x)$ and is not invertible. Then $$K[[x]] \otimes_{K[x]} K[x]/(f) \cong K[[x]]/(f) = 0$$ where the latter follows from the fact that since $f \notin (x)$, it becomes invertible in $K[[x]]$.

In particular, $K[[x]]$ is not faithfully flat over $K[x]$, so it, being a nonzero module, cannot be free over it.