I would need a result asserting that if, say, a locally free sheaf $\mathcal{F}$ on a projective $X$ has non-vanishing cohomology when restricted to any smooth hyperplane section, then $H^1(X, \mathcal{F}) \neq 0$, too. I fear that this is too naive, so I will be more specific in two steps, to contextualize. Also, to be fair, I should substitute "complex of sheaves" to "sheaf" (and "hypercohomology" to "cohomology").
Let $X$ be a projective complex surface, and $F^\bullet = F^1 \xrightarrow{\theta} F^2 \xrightarrow{\theta} F^3$ a 3-steps complex of locally free sheaves. Suppose that we know that for every smooth hyperplane section $i \colon C \subset X$, the hypercohomology $\mathbb{H}^1(C, i^*F^\bullet)$ of the restriction $i^*F^\bullet = i^*F^1 \xrightarrow{i^*\theta} i^*F^2$ is not zero (in my situation, I actually know that this has just 2 steps). Is it possible that $\mathbb{H}^1(X, F^\bullet) = 0$?
Some more background (explaining the, probably useless, condition on lengths of complexes). I actually have a flat bundle $\mathcal{V}$ on $X$, with a decomposition $\mathcal{V} = \bigoplus_k U_k$ and maps $\theta \colon U_k \to \mathcal{A}^{1,0}(U_{k+1})$, where $\mathcal{A}^{1,0}$ denotes $(1,0)$-forms, such that $\theta \wedge \theta = 0$, i.e., for every $k$ a complex $$ F_{X,k}^\bullet = U_k \xrightarrow{\theta} \mathcal{A}^{1,0}(U_{k+1}) \xrightarrow{\theta} \mathcal{A}^{2,0}(U_{k+2}). $$ By Lefschetz hyperplane theorem, since $\pi_1(C) \twoheadrightarrow \pi_1(X)$, we get an injection $H^1(X, \mathcal{V}) \hookrightarrow H^1(C, i^*\mathcal{V})$. The above decomposition gives a decomposition $$ H^1(X, \mathcal{V}) \cong \mathbb{H}^1\big(\mathcal{V} \xrightarrow{\theta}\mathcal{A^{1,0}(V)} \xrightarrow{\theta} \mathcal{A^{2,0}(V)}\big) = \bigoplus_k \mathbb{H}^1(F_{X,k}^\bullet), $$ and we actually have an analogous decomposition on $C$ (of course, with only 2 steps complexes, since on curves $\mathcal{A}^{2,0} = 0$). The natural injective map, then, sends $\mathbb{H}^1(F_{X,k}^\bullet)$ to $\mathbb{H}^1(F_{C,k}^\bullet)$, and every restriction must be injective. I would like to deduce that if the former vanishes for some $k$, then the latter does, too, for the same $k$ and at least one smooth hyperplane section $C$ (a fortiori, in my situation this will imply that it vanishes for every such $C$, but some other arguments are needed).
I apologize in advance for the long trivial-but-disguised-as-serious question, no matter how, basic algebraic geometry always escapes me...
I misread the question (in the opposite direction). The answer is yes, this can happen. Let $n\ge 1$ be an integer and let $F=O_X(-n)$. Let $C$ be any hyperplane section in $X$. We have $$ H^1(X, O_X(-n)|_C)=H^1(C, O_X(-n)|_C)\simeq H^0(C, \omega_C(n)).$$ The last term is not zero for $n$ big enough by Riemann-Roch. And the bound on $n$ depends only on the arithmetic genus of $C$, equal to $1-\chi(O_X)+\chi(O_X(-1))$. So for $n$ big enough, $H^1(X, O_X(-n)|_C)\ne 0$ for any hyperplane section $C$ (even singular, reducible etc).
On the other hand, at least when $X$ is Cohen-Macaulay (e.g. smooth), we have $$ H^1(X, O_X(-n))\simeq H^1(X, \omega_{X/k}(n))=0$$ for $n$ big enough by Serre vanishing.