= Solution
Since $X_0=0$, its <supremum> $X^\star$ is nonnegative. For $b>0$, continuity gives
$$
\{X^\star>b\}=\bigcup_{n\geq1}\{T_{b+1/n}<\infty\}.
$$
The events increase with $n$. Part (iii) therefore yields $\mathbb P(X^\star>b)=\lim_ne^{-(b+1/n)}=e^{-b}$. Letting $b\downarrow0$ gives $\mathbb P(X^\star>0)=1$; letting $b\to\infty$ gives finiteness <almost surely>. Consequently
$$
\boxed{X^\star\sim\operatorname{Exp}(1),\qquad
\mathbb P(X^\star\leq x)=
\begin{cases}0,&x<0,\\1-e^{-x},&x\geq0.\end{cases}}
$$
Its <probability density function> is $e^{-x}\mathbf1_{\{x>0\}}$, with no atom at zero. Using strict exceedances above also avoids presupposing that the infinite-horizon <supremum> is attained.
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