= Solution
Let $B$ be the selected block, and write $\alpha=\mathbb P(B=1)$. Because both the input and output <alphabets> are disjoint across blocks, $B$ is determined by either the input or the output. The mutual-information chain rule therefore gives
$$
I(X;Y)=H(B)+I(X;Y\mid B)\leq h_2(\alpha)+\alpha C_1+(1-\alpha)C_2.
$$
For any fixed $\alpha$, <equality> is achievable by using a capacity-achieving distribution inside each block. It remains to maximize this binary expression.
Put $S=2^{C_1}+2^{C_2}$ and $r=2^{C_1}/S$. A direct expansion of <binary relative entropy> yields
$$
h_2(\alpha)+\alpha C_1+(1-\alpha)C_2
=\log_2S-d\bigl((\alpha,1-\alpha),(r,1-r)\bigr).
$$
Part 1(a) makes the <maximum> $\log_2S$, uniquely at $\alpha=r$. Hence
$$
\boxed{C=\log_2(2^{C_1}+2^{C_2}),\qquad\mathbb P(B=i)=\frac{2^{C_i}}{2^{C_1}+2^{C_2}}.}
$$
The block choice itself communicates information, in addition to any information sent within the selected block.
This is <capacity of a disjoint sum of channels>.
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