Let $\mathrm{X}$ be a cubic $d$fold, and $\mathrm{F}(\mathrm{X})$ its Fano variety of lines. Is the integral cohomology of $\mathrm{F}(\mathrm{X})$ torsionfree? For $d=3$ A. Collino (`The fundamental group of the Fano surface, I') proves that there exists an exact sequence $$[\pi_1,\pi_1]\rightarrow\pi_1\rightarrow\mathbf{Z}^{\oplus 10}\rightarrow 0$$ where $\pi_1=\pi_1(\mathrm{F}(\mathrm{X}))$ (Prop. 2.3.4 in the paper). Hence $\mathrm{H}_1(\mathrm{F}(\mathrm{X});\mathbf{Z})=\pi_1^{\mathrm{ab}}$ is torsionfree, which is enough. For $d=4$ Beauville and Donagi show that $\mathrm{F}(\mathrm{X})$ is deformation equivalent to $\mathrm{Hilb}^2(\mathrm{K3})$, the integral cohomology of which is torsionfree (for example by the general results of this paper of Totaro). So the only remaining case is $d>4$, where $\mathrm{F}(\mathrm{X})$ is a Fano variety in the usual sense. Higherdimensional Fano varieties can have torsion in their integral cohomology, see for example here.

3$\begingroup$ According to our work with Galkin, arxiv.org/pdf/1405.5154.pdf, to show that $H^*(F(X), \mathbb{Z})$ are torsionfree it suffices to show that $H^*(Hilb_2(X), \mathbb{Z})$ are torsion free. Indeed this follows from Thm 5.1 by taking the realization in the Grothendieck ring of integral Hodge structures (we do not do this in the paper; for rational Hodge structures this is explained on page 8 and goes back to Bittner). Now if $X$ is evendimensional, Totaro's result in the post implies that this indeed is the case since cohomology of $X$ are even and torsionfree. $\endgroup$ – Evgeny Shinder Oct 13 '19 at 19:58

3$\begingroup$ Actually, the argument above seems to be true for all cubics (odd or evendimensional), as according to Totaro's paper arxiv.org/pdf/1506.00968.pdf, Thm 2.2, $Hilb_2(X)$ has torsion free cohomology as soon as this holds for $X$; in particular it applies to any hypersurface. $\endgroup$ – Evgeny Shinder Oct 13 '19 at 22:02
Here is an expanded version of my comments. Let's work over the complex numbers which I suppose is assumed in the question. Let $K_0(Var)$ be the Grothendieck ring of varieties (see e.g. https://arxiv.org/abs/1405.5154). Furthemore, let $K_0(HS)$ be the Grothendieck group of integral polarizable Hodge structures with respect to direct sum operation, and let $K_0(FinAb)$ be the Grothendieck group of finite abelian groups also with respect to direct sum operation.
We have a composition of group homomorphisms $$ K_0(Var) \to K_0(HS) \to K_0(FinAb) $$ where the first map takes a class of a smooth projective variety to its total Hodge structure, and the second map takes a Hodge structure to its torsion part (forgetting the weights). Section 3 in Efimov's paper https://arxiv.org/pdf/1707.08997.pdf gives the details for the definition of the first map, from Bittner's blow up presentation of the Grothendieck ring of varieties. Second map is straightforward. Existence of these maps implies that classes of smooth projective varieties control the torsion in cohomology groups.
Now let $X$ be a smooth cubic of dimension $n$. According to https://arxiv.org/abs/1405.5154, Thm. 5.1, the class of the Fano variety $F(X)$ in the Grothendieck ring is expressed as a combination of $Hilb_2(X)$ and $\mathbb{P}^n \times X$ (with a coefficient $\mathbb{L}^2$ that shifts cohomology groups). Explicitly in $K_0(FinAb)$ we have the following: $$ [H^*(F(X), \mathbb{Z})_{tors}] + (n+1) [H^*(X, \mathbb{Z})_{tors}] = [H^*(Hilb_2(X), \mathbb{Z})_{tors}] $$ In fact, since the category of finite abelian groups has unique decompositions into indecomposables, equality of classes implies isomorphism hence $$ H^*(F(X), \mathbb{Z})_{tors} \oplus H^*(X, \mathbb{Z})_{tors}^{\oplus(n+1)} \simeq H^*(Hilb_2(X), \mathbb{Z})_{tors} $$
Thus it suffices to show that $Hilb_2(X)$ has torsionfree cohomology groups. All hypersurfaces are torsionfree, and by Totaro's result https://arxiv.org/pdf/1506.00968.pdf, Thm 2.2, same is true for $Hilb_2(X)$.