= Solution
The <polynomial ring> $\mathbb F_2[X]$ is a <euclidean domain> under <polynomial degree>, hence a <principal ideal domain>. By part (b)(i), each
$$
K_j=\mathbb F_2[X]/(f_j)
$$
is a <field>. By <polynomial division>, every residue class has exactly one representative of <polynomial degree> less than $n_j$. Thus $1,\overline X,\ldots,\overline X^{n_j-1}$ form a <basis> over $\mathbb F_2$, and $|K_j|=2^{n_j}$.
For $j\ne k$, the <ideals> $(f_j)$ and $(f_k)$ are distinct <maximal ideals>, so they are <comaximal ideals>: their sum is an <ideal> properly containing $(f_j)$ and must be the whole <ring>. The <Chinese remainder theorem for ideals> and part (b)(ii) therefore give the <ring isomorphism>
$$
\boxed{\mathbb F_2[X]/(f_1\cdots f_r)\ \cong\ \prod_{j=1}^rK_j,\qquad\overline g\longmapsto(g+(f_j))_{j=1}^r}.
$$
In particular, this is a product of <finite fields> of the required sizes. The map preserves multiplication as well as addition and is surjective by the <Chinese remainder theorem for ideals>.
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