Smoothness of Picard scheme I am reading something about classification od minimal model of rational surfaces.
And there is a case when the Picard number of a smooth projective surface $X$ is 1 and anticanonical bundle is ample. Then it said that the its Picard scheme’s component $Pic^0_X$ is smooth. How could we get this?

I don't think the condition on the Picard number is required. A variety with ample anticanonical bundle is called Fano variety. In the case of surfaces, one also speaks about del Pezzo surfaces. For any smooth Fano variety one has $H^1(X, \mathcal{O}_X)=0$. In characteristic zero, this follows from Kodaira vanishing. In characteristic $p > 0$, general Kodaira vanishing does not hold (not even for del Pezzo surfaces), but $H^1(X, \mathcal{O}_X) = 0$ is still true. See for example Thm1.1 in Fano threefolds in positive characteristic, Shepherd Barron.
But $h^1(X, \mathcal{O}_X) = \mathrm{dim}T_0(\mathrm{Pic}_X)$. Therefore, $\mathrm{Pic}^0(X)$ is just a disjoint union of points and thus smooth. 
