Part (1) is immediate from property (2) for siblings. \medskip\noindent Assume $F$ is fully faithful. Denote $\mathcal{D}' \subset \mathcal{D}$ the essential image of $F$ so that $F : D^b(\mathcal{A}) \to \mathcal{D}'$ is an equivalence. Since the functor $F'$ factors through $\mathcal{D}'$ by property (2) for siblings, we can consider the functor $H = F^{-1} \circ F' : D^b(\mathcal{A}) \to D^b(\mathcal{A})$. Observe that $H$ is a sibling of the identity functor. Since it suffices to prove that $H$ is fully faithful, we reduce to the problem discussed in the next paragraph. \medskip\noindent Set $\mathcal{D} = D^b(\mathcal{A})$. We have to show a sibling $F : \mathcal{D} \to \mathcal{D}$ of the identity functor is fully faithful. Denote $a_X : X \to F(X)$ the functorial isomorphism for $X \in \Ob(\mathcal{A})$ given to us by Definition \ref{definition-siblings}. For any $K$ in $\mathcal{D}$ and distinguished triangle $K_1 \to K_2 \to K_3$ of $\mathcal{D}$ if the maps $$ F : \Hom(K, K_i[n]) \to \Hom(F(K), F(K_i[n])) $$ are isomorphisms for all $n \in \mathbf{Z}$ and $i = 1, 3$, then the same is true for $i = 2$ and all $n \in \mathbf{Z}$. This uses the $5$-lemma Homology, Lemma \ref{homology-lemma-five-lemma} and Derived Categories, Lemma \ref{derived-lemma-representable-homological}; details omitted. Similarly, if the maps $$ F : \Hom(K_i[n], K) \to \Hom(F(K_i[n]), F(K)) $$ are isomorphisms for all $n \in \mathbf{Z}$ and $i = 1, 3$, then the same is true for $i = 2$ and all $n \in \mathbf{Z}$. Using the canonical truncations and induction on the number of nonzero cohomology objects, we see that it is enough to show $$ F : \Ext^q(X, Y) \to \Ext^q(F(X), F(Y)) $$ is bijective for all $X, Y \in \Ob(\mathcal{A})$ and all $q \in \mathbf{Z}$. Since $F$ is a sibling of $\text{id}$ we have $F(X) \cong X$ and $F(Y) \cong Y$ hence the right hand side is zero for $q < 0$. The case $q = 0$ is OK by our assumption that $F$ is a sibling of the identity functor. It remains to prove the cases $q > 0$. \medskip\noindent The case $q = 1$: Injectivity. An element $\xi$ of $\Ext^1(X, Y)$ gives rise to a distinguished triangle $$ Y \to E \to X \xrightarrow{\xi} Y[1] $$ Observe that $E \in \Ob(\mathcal{A})$. Since $F$ is a sibling of the identity functor we obtain a commutative diagram $$ \xymatrix{ E \ar[d] \ar[r] & X \ar[d] \\ F(E) \ar[r] & F(X) } $$ whose vertical arrows are the isomorphisms $a_E$ and $a_X$. By TR3 the distinguished triangle associated to $\xi$ we started with is isomorphic to the distinguished triangle $$ F(Y) \to F(E) \to F(X) \xrightarrow{F(\xi)} F(Y[1]) = F(Y)[1] $$ Thus $\xi = 0$ if and only if $F(\xi)$ is zero, i.e., we see that $F : \Ext^1(X, Y) \to \Ext^1(F(X), F(Y))$ is injective. \medskip\noindent The case $q = 1$: Surjectivity. Let $\theta$ be an element of $\Ext^1(F(X), F(Y))$. This defines an extension of $F(X)$ by $F(Y)$ in $\mathcal{A}$ which we may write as $F(E)$ as $F$ is a sibling of the identity functor. We thus get a distinguished triangle $$ F(Y) \xrightarrow{F(\alpha)} F(E) \xrightarrow{F(\beta)} F(X) \xrightarrow{\theta} F(Y[1]) = F(Y)[1] $$ for some morphisms $\alpha : Y \to E$ and $\beta : E \to X$. Since $F$ is a sibling of the identity functor, the sequence $0 \to Y \to E \to X \to 0$ is a short exact sequence in $\mathcal{A}$! Hence we obtain a distinguished triangle $$ Y \xrightarrow{\alpha} E \xrightarrow{\beta} X \xrightarrow{\delta} Y[1] $$ for some morphism $\delta : X \to Y[1]$. Applying the exact functor $F$ we obtain the distinguished triangle $$ F(Y) \xrightarrow{F(\alpha)} F(E) \xrightarrow{F(\beta)} F(X) \xrightarrow{F(\delta)} F(Y)[1] $$ Arguing as above, we see that these triangles are isomorphic. Hence there exists a commutative diagram $$ \xymatrix{ F(X) \ar[d]^\gamma \ar[r]_{F(\delta)} & F(Y[1]) \ar[d]_\epsilon \\ F(X) \ar[r]^\theta & F(Y[1]) } $$ for some isomorphisms $\gamma$, $\epsilon$ (we can say more but we won't need more information). We may write $\gamma = F(\gamma')$ and $\epsilon = F(\epsilon')$. Then we have $\theta = F(\epsilon' \circ \delta \circ (\gamma')^{-1})$ and we see the surjectivity holds. \medskip\noindent The case $q > 1$: surjectivity. Using Yoneda extensions, see Derived Categories, Section \ref{derived-section-ext}, we find that for any element $\xi$ in $\Ext^q(F(X), F(Y))$ we can find $F(X) = B_0, B_1, \ldots, B_{q - 1}, B_q = F(Y) \in \Ob(\mathcal{A})$ and elements $$ \xi_i \in \Ext^1(B_{i - 1}, B_i) $$ such that $\xi$ is the composition $\xi_q \circ \ldots \circ \xi_1$. Write $B_i = F(A_i)$ (of course we have $A_i = B_i$ but we don't need to use this) so that $$ \xi_i = F(\eta_i) \in \Ext^1(F(A_{i - 1}), F(A_i)) \quad\text{with}\quad \eta_i \in \Ext^1(A_{i - 1}, A_i) $$ by surjectivity for $q = 1$. Then $\eta = \eta_q \circ \ldots \circ \eta_1$ is an element of $\Ext^q(X, Y)$ with $F(\eta) = \xi$. \medskip\noindent The case $q > 1$: injectivity. An element $\xi$ of $\Ext^q(X, Y)$ gives rise to a distinguished triangle $$ Y[q - 1] \to E \to X \xrightarrow{\xi} Y[q] $$ Applying $F$ we obtain a distinguished triangle $$ F(Y)[q - 1] \to F(E) \to F(X) \xrightarrow{F(\xi)} F(Y)[q] $$ If $F(\xi) = 0$, then $F(E) \cong F(Y)[q - 1] \oplus F(X)$ in $\mathcal{D}$, see Derived Categories, Lemma \ref{derived-lemma-split}. Since $F$ is a sibling of the identity functor we have $E \cong F(E)$ and hence $$ E \cong F(E) \cong F(Y)[q - 1] \oplus F(X) \cong Y[q - 1] \oplus X $$ In other words, $E$ is isomorphic to the direct sum of its cohomology objects. This implies that the initial distinguished triangle is split, i.e., $\xi = 0$.