= Solution
Write
$$
\pi(r)=\mathbb P_{1/2}(0\longleftrightarrow\partial\Lambda_r).
$$
The <Russo-Seymour-Welsh theorem> and the <Harris-FKG inequality> give the standard <one-arm extension estimate>: there is $c>0$ such that
$$
\pi(2r)\geq c\pi(r)
$$
uniformly in $r$. Indeed, on the one-arm event to scale $r$, a fixed finite collection of open rectangle crossings in the annulus $\Lambda_{2r}\setminus\Lambda_r$, each having probability bounded below by RSW, joins that arm to $\partial\Lambda_{2r}$; FKG multiplies the lower bounds. Iteration shows that $\pi(ar)$ and $\pi(r)$ are comparable for every fixed $a>0$. This proves the estimate suggested in the hint.
Fix $x\in\partial\Lambda_n$ and choose $r=\lfloor n/4\rfloor$. If $0\longleftrightarrow x$, there is an open arm from $0$ to distance $r$ and another from $x$ to distance $r$. These are <independent events> because they use disjoint edge sets, so the extension estimate gives
$$
\mathbb P_{1/2}(0\longleftrightarrow x)
\leq\pi(r)^2\leq C\pi(n)^2.
$$
For the reverse inequality, take one-arm events from $0$ and $x$ at scale comparable with $n$, in disjoint boxes. A fixed collection of open crossings of rectangles of bounded aspect ratio joins the two arms. The <Russo-Seymour-Welsh theorem> bounds the probability of every added crossing below uniformly in $n$ and in the position of $x$ along the four sides; the <Harris-FKG inequality> and the arm-extension estimate therefore give
$$
\mathbb P_{1/2}(0\longleftrightarrow x)
\geq c\pi(n)^2.
$$
This is the usual <RSW gluing lemma for two one-arm events>. Enlarging the constants handles the finitely many small $n$, proving the claim with positive constants $c_1,c_2$.
Solved by gpt-5.6-sol high.
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