= Solution
First suppose $0<r<\rho=|z_0|$. For every $R>\rho$, the <planar Brownian annulus hitting probability> gives
$$
\mathbb P(\tau_r<\infty)
\geq\mathbb P(\tau_r<\tau_R)
=\frac{\log(R/\rho)}{\log(R/r)}\longrightarrow1
\quad(R\to\infty).
$$
Hence $\tau_r<\infty$ <almost surely>. This uses increasing outer radii; letting the inner radius tend to zero instead proves the different claim in part (ii).
For $r>\rho$, the disk-exit estimate in (i), with outer radius $r$, directly yields $\mathbb E\tau_r\leq(r^2-\rho^2)/2<\infty$.
For $r=\rho$, retain the definition $\tau_r=\inf\{t>0:|B_t|=r\}$. The first case shows that radius $\rho/2$ is reached in finite time. Starting there, the <Strong Markov property> of <Brownian motion> and the disk-exit argument show that radius $\rho$ is subsequently reached in finite time. Thus the infimum over positive hitting times is finite in this case as well. Altogether,
$$
\boxed{\mathbb P(\tau_r<\infty)=1\qquad\text{for every }r>0.}
$$
The conclusion can also be made simultaneous in all $r>0$: intersect the probability-one events for positive rational radii, then use continuity to cross each intermediate radius.
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