= Solution
Let $F_n$ be the <distribution function> of $X_n$. The limiting constant zero has <distribution function> $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\geq0$. For every $\varepsilon>0$, both $-\varepsilon$ and $\varepsilon$ are continuity points of $F$. Thus <convergence in distribution> implies $F_n(-\varepsilon)\to0$ and $F_n(\varepsilon)\to1$. Therefore
$$
0\leq\mathbb P(|X_n|>\varepsilon)
\leq F_n(-\varepsilon)+1-F_n(\varepsilon)\longrightarrow0.
$$
This proves
$$
\boxed{X_n\xrightarrow d0\ \Longrightarrow\ X_n\xrightarrow{\mathbb P}0.}
$$
The same proof after subtracting a fixed real $c$ proves <convergence in distribution to a constant implies convergence in probability>. A deterministic limit fixes the coupling automatically; the counterexample in (c) exploits a nonconstant limit whose <probability distribution> alone does not fix that coupling.
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