= Solution
Identify a binary word with its coefficient <polynomial> in $R=\mathbb F_2[X]/(X^7-1)$. A <cyclic code> is exactly an <ideal> of $R$: closure under multiplication by $X$ is closure under cyclic shift. Its inverse image in the <principal ideal domain> $\mathbb F_2[X]$ has a unique monic <generator polynomial of a cyclic code> $g$ dividing $X^7-1$. Put
$$
a=X+1,\quad b=X^3+X+1,\quad c=X^3+X^2+1;\qquad X^7-1=abc.
$$
The cubics have no root in $\mathbb F_2$, hence are <irreducible polynomials>. The eight monic <divisors> yield all <binary cyclic codes of length seven>. A code generated by $g$ has <basis> $g,Xg,\ldots,X^{6-\deg g}g$ and <dimension> $7-\deg g$, with the zero code treated separately.
For the <dual of a cyclic code>, let $h=(X^7-1)/g$. Orthogonality of coefficient <vectors> to all cyclic shifts gives the generator $h^*(X)=X^{\deg h}h(X^{-1})$. Here $a^*=a,b^*=c,c^*=b$. Thus the complete pairing, with the respective <dimensions>, is
$$
\boxed{1\ (7)\longleftrightarrow abc\ (0),\quad
a\ (6)\longleftrightarrow bc\ (1),\quad
b\ (4)\longleftrightarrow ab\ (3),\quad
c\ (4)\longleftrightarrow ac\ (3).}
$$
These are respectively the whole-space/zero, even-weight/repetition, and two <Hamming code>/<binary simplex code> pairs. The reciprocal formula follows because the product of a word <polynomial> and the reversed <polynomial> of an orthogonal word has zero coefficients in all cyclic positions; the resulting inclusion has the correct complementary <dimension>, hence is equality.
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