To see part (1) let $y$ be an object of $\mathcal{Y}$ which lies over a scheme $V$ such that the corresponding morphism $y : V \to \mathcal{Y}$ is flat. Then $g(y) : V \to \mathcal{Y} \to \mathcal{X}$ is flat as a composition of flat morphisms (see Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-composition-flat}) hence $\mathcal{F}(g(y))$ is zero by assumption. Since $g^*\mathcal{F} = g^{-1}\mathcal{F}(y) = \mathcal{F}(g(y))$ we conclude $g^*\mathcal{F}$ is parasitic. \medskip\noindent To see part (2)(a) note that if $\{x_i \to x\}$ is a $\tau$-covering of $\mathcal{X}$, then each of the morphisms $x_i \to x$ lies over a flat morphism of schemes. Hence if $x$ lies over a scheme $U$ such that $x : U \to \mathcal{X}$ is flat, so do all of the objects $x_i$. Hence the presheaf $\mathcal{F}^+$ (see Sites, Section \ref{sites-section-sheafification}) is parasitic if the presheaf $\mathcal{F}$ is parasitic. This proves (2)(a) as the sheafification of $\mathcal{F}$ is $(\mathcal{F}^+)^+$. \medskip\noindent Let $\mathcal{F}$ be a parasitic $\tau$-module. It is immediate from the definitions that any submodule of $\mathcal{F}$ is parasitic. On the other hand, if $\mathcal{F}' \subset \mathcal{F}$ is a submodule, then it is equally clear that the presheaf $x \mapsto \mathcal{F}(x)/\mathcal{F}'(x)$ is parasitic. Hence the quotient $\mathcal{F}/\mathcal{F}'$ is a parasitic module by (2)(a). Finally, we have to show that given a short exact sequence $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ with $\mathcal{F}_1$ and $\mathcal{F}_3$ parasitic, then $\mathcal{F}_2$ is parasitic. This follows immediately on evaluating on $x$ lying over a scheme flat over $\mathcal{X}$. This proves (2)(b), see Homology, Lemma \ref{homology-lemma-characterize-serre-subcategory}. \medskip\noindent Let $f_i : \mathcal{X}_i \to \mathcal{X}$ be a jointly surjective family of smooth morphisms of algebraic stacks and assume each $f_i^*\mathcal{F}$ is parasitic. Let $x$ be an object of $\mathcal{X}$ which lies over a scheme $U$ such that $x : U \to \mathcal{X}$ is flat. Consider a surjective smooth covering $W_i \to U \times_{x, \mathcal{X}} \mathcal{X}_i$. Denote $y_i : W_i \to \mathcal{X}_i$ the projection. It follows that $\{f_i(y_i) \to x\}$ is a covering for the smooth topology on $\mathcal{X}$. Since a composition of flat morphisms is flat we see that $f_i^*\mathcal{F}(y_i) = 0$. On the other hand, as we saw in the proof of (1), we have $f_i^*\mathcal{F}(y_i) = \mathcal{F}(f_i(y_i))$. Hence we see that for some smooth covering $\{x_i \to x\}_{i \in I}$ in $\mathcal{X}$ we have $\mathcal{F}(x_i) = 0$. This implies $\mathcal{F}(x) = 0$ because the smooth topology is the same as the \'etale topology, see More on Morphisms, Lemma \ref{more-morphisms-lemma-etale-dominates-smooth}. Namely, $\{x_i \to x\}_{i \in I}$ lies over a smooth covering $\{U_i \to U\}_{i \in I}$ of schemes. By the lemma just referenced there exists an \'etale covering $\{V_j \to U\}_{j \in J}$ which refines $\{U_i \to U\}_{i \in I}$. Denote $x'_j = x|_{V_j}$. Then $\{x'_j \to x\}$ is an \'etale covering in $\mathcal{X}$ refining $\{x_i \to x\}_{i \in I}$. This means the map $\mathcal{F}(x) \to \prod_{j \in J} \mathcal{F}(x'_j)$, which is injective as $\mathcal{F}$ is a sheaf in the \'etale topology, factors through $\mathcal{F}(x) \to \prod_{i \in I} \mathcal{F}(x_i)$ which is zero. Hence $\mathcal{F}(x) = 0$ as desired. \medskip\noindent Proof of (4): omitted. Hint: similar, but simpler, than the proof of (3).