= Solution
Under the bounded-support convention, \b[false in general, true if $f(0)=0$]. For the first claim, take $g=0$ and $f\equiv1$: the composition is one on all of $\mathbb R$, and cannot have bounded support. For the second, refine the finitely many defining intervals into disjoint intervals on which $g$ is constant. The composition is constant on each of them and vanishes outside their bounded union because $f(0)=0$. Hence it is another <step function>.
Under the unrestricted piecewise-constant convention, \b[true in both cases]: composition with any function preserves constancy on the same partition, including its unbounded cells. This also applies to the broader measurable-partition convention. <Continuity> of $f$ is not needed for this particular assertion.
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