= Solution
The <standard deviations> defining the position and momentum uncertainties satisfy $(\Delta x)^2=\langle x^2\rangle-\langle x\rangle^2$ and $(\Delta p)^2=\langle p^2\rangle-\langle p\rangle^2$. Part (a) makes both means zero. Using the second moments from part (b),
$$
\Delta x=\sqrt{\frac{\hbar(n+1/2)}{m\omega}},\qquad
\Delta p=\sqrt{m\hbar\omega(n+1/2)}.
$$
Multiplication gives
$$
\boxed{\Delta x\,\Delta p=\hbar\left(n+\frac12\right).}
$$
The <Heisenberg uncertainty relation> requires only $\Delta x\Delta p\geq\hbar/2$. Among the oscillator <energy eigenstates>, equality occurs solely for the <ground state> $n=0$; the excited states have larger uncertainty products.
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