Proof.
Assume (1). We may assume $K$ is the complex
\[ \ldots \to 0 \to P^ a \to P^{a + 1} \to \ldots \to P^{b - 1} \to P^ b \to 0 \to \ldots \]
where $P^ i$ is a projective $R$-module for all $i \in \mathbf{Z}$. In this case we can compute the ext groups by the complex
\[ \ldots \to 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(P^ b, N) \to \ldots \to \mathop{\mathrm{Hom}}\nolimits _ R(P^ a, N) \to 0 \to \ldots \]
and we obtain (2).
Assume (2) holds. Choose an injection $H^ n(K) \to I$ where $I$ is an injective $R$-module. Since $\mathop{\mathrm{Hom}}\nolimits _ R(-, I)$ is an exact functor, we see that $\mathop{\mathrm{Ext}}\nolimits ^{-n}(K, I) = \mathop{\mathrm{Hom}}\nolimits _ R(H^ n(K), I)$. We conclude in particular that $H^ n(K)$ is zero for $n > b$. Thus (2) implies (3).
By the same argument as in (2) implies (3) gives that (3) implies (4).
Assume (4). The same argument as in (2) implies (3) shows that $H^{a - 1}(K) = 0$, i.e., we have $H^ i(K) = 0$ unless $i \in [a, b]$. In particular, $K$ is bounded above and we can choose a a complex $P^\bullet $ representing $K$ with $P^ i$ projective (for example free) for all $i \in \mathbf{Z}$ and $P^ i = 0$ for $i > b$. See Derived Categories, Lemma 13.15.4. Let $Q = \mathop{\mathrm{Coker}}(P^{a - 1} \to P^ a)$. Then $K$ is quasi-isomorphic to the complex
\[ \ldots \to 0 \to Q \to P^{a + 1} \to \ldots \to P^ b \to 0 \to \ldots \]
as $H^ i(K) = 0$ for $i < a$. Denote $K' = (P^{a + 1} \to \ldots \to P^ b)$ the corresponding object of $D(R)$. We obtain a distinguished triangle
\[ K' \to K \to Q[-a] \to K'[1] \]
in $D(R)$. Thus for every $R$-module $N$ an exact sequence
\[ \mathop{\mathrm{Ext}}\nolimits ^{-a}(K', N) \to \mathop{\mathrm{Ext}}\nolimits ^1(Q, N) \to \mathop{\mathrm{Ext}}\nolimits ^{1 - a}(K, N) \]
By assumption the term on the right vanishes. By the implication (1) $\Rightarrow $ (2) the term on the left vanishes. Thus $Q$ is a projective $R$-module by Algebra, Lemma 10.77.2. Hence (1) holds and the proof is complete.
$\square$
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