Yes, all of this configurations do exist provably in $\mathsf{ZFC}$.
Let us construct $\alpha<\beta$ such that $(L_{\alpha+n};\in,\alpha,\langle c\mid c\in L_\alpha\rangle)\equiv(L_{\beta+n};\in,\beta,\langle c\mid c\in L_\alpha\rangle)$. Consider the structure $(L_{\omega_1+n};\in,\omega_1)$, where the last $\omega_1$ is a constant. Next we construct a sequence of countable elementary substructures of this structure $$\mathfrak{M}_0\prec \mathfrak{M}_1\prec \ldots,$$ such that $\mathfrak{M}_{i+1}\supseteq L_{\delta +1}$, for all $\delta<\omega_1$ for which $\mathfrak{M}_i\cap (L_{\delta+1}\setminus L_\delta)\ne \emptyset$. Let $\mathfrak{M}_{\omega}=\bigcup_{i<\omega} \mathfrak{M}_i$. By Condensation Lemma, $\mathfrak{M}_\omega$ is isomorphic to a structure of the form $(L_\gamma;\in,\alpha)$, where $\gamma$ is countable and $\alpha<\gamma$. Furthermore by construction of $\mathfrak{M}_\omega$, $L_\alpha\supseteq \mathfrak{M}_\omega$ and the isomorphism keeps all the elements of $L_\alpha$ in place. Also using the elementary equivalence $(L_\gamma;\in,\alpha)\equiv(L_{\omega_1+n};\in,\omega_1)$ it is easy to see that $\gamma=\alpha+n$.
In the same manner we construct countable $\gamma$ s.t. $(L_{\gamma};\in,\alpha,\langle c\mid c\in L_\alpha\rangle)\equiv(L_{\omega_12};\in,\omega_1,\langle c\mid c\in L_\alpha\rangle)$ and again using the elementary equivalence we show that $\gamma=\alpha2$.
Given any ordinal $\delta$, we consider $(L_{\omega_{\delta+1}+\delta};\in,\omega_{\delta +1},\delta)$ and construct its elementary submodel of the cardinality $\le \max(|\delta|,\aleph_0)$ that will contain all ordinals $\le \delta$ and have the property that for any $\eta<\omega_{\delta+1}$, whenever some element of $L_{\eta+1}\setminus L_\eta$ is in the submodel, then whole $L_{\eta+1}$ is contained in the submodel. This gives us $\delta<\alpha<\gamma< \omega_{\delta+1}$ s.t. $(L_{\gamma};\in,\alpha,\langle c\mid c\in L_\alpha\rangle)\equiv(L_{\omega_{\delta+1}+\delta};\in,\omega_{\delta +1},\langle c\mid c\in L_\alpha\rangle)$. Again using the elementary equivalence we see that $\gamma=\alpha+\delta$.