Proof of (1). We will use without further mention that finite limits and colimits of constructible sheaves are constructible (Lemma \ref{lemma-kernel-finite-locally-constant}). For each $i$ let $T_i \subset X$ be the set of points $x \in X$ where $\mathcal{F}_{i, \overline{x}} \to \mathcal{F}_{\overline{x}}$ is not surjective. Because $\mathcal{F}_i$ and $\mathcal{F}$ are constructible $T_i$ is a constructible subset of $X$ (Lemma \ref{lemma-support-constructible}). Since the stalks of $\mathcal{F}$ are finite and since $\mathcal{F} = \colim_{i \in I} \mathcal{F}_i$ we see that for all $x \in X$ we have $x \not \in T_i$ for $i$ large enough. Since $X$ is a spectral space by Properties, Lemma \ref{properties-lemma-quasi-compact-quasi-separated-spectral} the constructible topology on $X$ is quasi-compact by Topology, Lemma \ref{topology-lemma-constructible-hausdorff-quasi-compact}. Thus $T_i = \emptyset$ for $i$ large enough. Thus $\mathcal{F}_i \to \mathcal{F}$ is surjective for $i$ large enough. Assume now that $\mathcal{F}_i \to \mathcal{F}$ is surjective for all $i$. Choose $i \in I$. For $i' \geq i$ denote $S_{i'} \subset X$ the set of points $x$ such that the number of elements in $\Im(\mathcal{F}_{i, \overline{x}} \to \mathcal{F}_{\overline{x}})$ is equal to the number of elements in $\Im(\mathcal{F}_{i, \overline{x}} \to \mathcal{F}_{i', \overline{x}})$. Because $\mathcal{F}_i$, $\mathcal{F}_{i'}$ and $\mathcal{F}$ are constructible $S_{i'}$ is a constructible subset of $X$ (details omitted; hint: use Lemma \ref{lemma-support-constructible}). Since the stalks of $\mathcal{F}_i$ and $\mathcal{F}$ are finite and since $\mathcal{F} = \colim_{i' \geq i} \mathcal{F}_{i'}$ we see that for all $x \in X$ we have $x \not \in S_{i'}$ for $i'$ large enough. By the same argument as above we can find a large $i'$ such that $S_{i'} = \emptyset$. Thus $\mathcal{F}_i \to \mathcal{F}_{i'}$ factors through $\mathcal{F}$ as desired. \medskip\noindent Proof of (2). Observe that a constructible abelian sheaf is a constructible sheaf of sets. Thus case (2) follows from (1). \medskip\noindent Proof of (3). We will use without further mention that the category of constructible sheaves of $\Lambda$-modules is abelian (Lemma \ref{lemma-kernel-finite-locally-constant}). For each $i$ let $\mathcal{Q}_i$ be the cokernel of the map $\mathcal{F}_i \to \mathcal{F}$. The support $T_i$ of $\mathcal{Q}_i$ is a constructible subset of $X$ as $\mathcal{Q}_i$ is constructible (Lemma \ref{lemma-support-constructible}). Since the stalks of $\mathcal{F}$ are finite $\Lambda$-modules and since $\mathcal{F} = \colim_{i \in I} \mathcal{F}_i$ we see that for all $x \in X$ we have $x \not \in T_i$ for $i$ large enough. Since $X$ is a spectral space by Properties, Lemma \ref{properties-lemma-quasi-compact-quasi-separated-spectral} the constructible topology on $X$ is quasi-compact by Topology, Lemma \ref{topology-lemma-constructible-hausdorff-quasi-compact}. Thus $T_i = \emptyset$ for $i$ large enough. This proves the first assertion. For the second, assume now that $\mathcal{F}_i \to \mathcal{F}$ is surjective for all $i$. Choose $i \in I$. For $i' \geq i$ denote $\mathcal{K}_{i'}$ the image of $\Ker(\mathcal{F}_i \to \mathcal{F})$ in $\mathcal{F}_{i'}$. The support $S_{i'}$ of $\mathcal{K}_{i'}$ is a constructible subset of $X$ as $\mathcal{K}_{i'}$ is constructible. Since the stalks of $\Ker(\mathcal{F}_i \to \mathcal{F})$ are finite $\Lambda$-modules and since $\mathcal{F} = \colim_{i' \geq i} \mathcal{F}_{i'}$ we see that for all $x \in X$ we have $x \not \in S_{i'}$ for $i'$ large enough. By the same argument as above we can find a large $i'$ such that $S_{i'} = \emptyset$. Thus $\mathcal{F}_i \to \mathcal{F}_{i'}$ factors through $\mathcal{F}$ as desired.