$S1:$ Yes. Since the rows of $A$ are linearly independent, $rank(A) = m$. So the column space of $A$ is an $m$-dimensional subspace of $\mathbb{R}^m$, i.e., is $\mathbb{R}^3$ itself. It follows that for any $b$, the equation $Ax = b$ is always solvable.
$S2:$ No, the solution maybe not unique. Since $rank(A) = m$, the nullspace is $n-m$ dimensional. Thus the solution is not unique if $n > m$, and is unique if $n = m$. (It will never have that $n < m$, otherwise the rows are not linearly independent.)
