= Solution
Let $B_n=\phi(0\leftrightarrow\partial\Lambda_n)$ and $s_n=|\partial\Lambda_n|$. This connection event is the union over $x\in\partial\Lambda_n$ of $\{0\leftrightarrow x\}$. The <union bound> therefore supplies a <graph vertex> with connection <probability> at least $B_n/s_n$.
Every boundary <graph vertex> has some coordinate equal to $n$ or $-n$. By applying a <permutation> and a <graph automorphisms>, which preserve both the law and the boundary, choose such a <graph vertex> in the form $x=(n,x_2,\ldots,x_d)$ while retaining its <probability> bound. Now $e_{2n}-x=(n,-x_2,\ldots,-x_d)$ is a reflection of $x$. Translation and reflection invariance give
$$
\phi(x\leftrightarrow e_{2n})=\phi(0\leftrightarrow e_{2n}-x)=\phi(0\leftrightarrow x).
$$
<Positive association of random variables> applied to the two connection events, whose intersection implies $0\leftrightarrow e_{2n}$, proves the <reflection comparison of random-cluster connections>:
$$
\boxed{C_{2n}\ge\phi(0\leftrightarrow x)^2,\qquad
\phi(0\leftrightarrow x)\ge B_n/s_n.}
$$
Also $C_n\le B_n$, since any path to $e_n$ meets the box boundary. Consequently
$$
C_n\le B_n\le s_n\sqrt{C_{2n}},
$$
and hence
$$
-\frac1{2n}\log C_{2n}-\frac{\log s_n}{n}
\le-\frac1n\log B_n
\le-\frac1n\log C_n.
$$
Here $s_n=(2n+1)^d-(2n-1)^d=O(n^{d-1})$, so $\log s_n/n\to0$. Both outside logarithmic connection rates converge to $\alpha(p,q)$ by part (b). Squeezing gives the rate for every box size, and therefore for the requested even subsequence:
$$
\boxed{-\frac1{2n}\log\phi^1_{p,q}(0\leftrightarrow\partial\Lambda_{2n})\longrightarrow\alpha(p,q).}
$$
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