Let $X$ be a normal, projective (complex) variety with at worst rational singularities. Let $\pi:Y \to X$ be the resolution of singularities obtained by blowing-up the singular points. Is $R^1 \pi\_\*\mathbb{Z}=0$? I am mainly interested in the case when $X$ is of dimension $3$.



Higher direct image for rational singularities
This should be true. Use the exponential sequence to obtain
$$\pi\_\*\mathcal{O}\_Y\to \pi\_\*\mathcal{O}\_Y^\*\to R^1\pi\_\*\mathbb{Z}\to R^1\pi\_\*\mathcal{O}\_Y$$
The last term is zero since you have rational singularities. You can check that the first map is surjective as follows: By normality
$$\pi\_\*\mathcal{O}\_Y = \mathcal{O}\_X,\quad \pi\_\*\mathcal{O}\_Y^\* = \mathcal{O}\_X^\*$$
Choose a local embedding $X\subset Z$, with $Z$ smooth. Consider the diagram
$$\begin{array}{ccc}
 \mathcal{O}\_Z & \xrightarrow{exp}&\mathcal{O}\_Z^\* \\ 
\downarrow&&\downarrow\\
 \mathcal{O}\_X & \xrightarrow{exp}&\mathcal{O}\_X^\*
\end{array}
$$
The top arrow $exp$ is surjective, and the vertical arrows are surjective. Therefore the bottom $\exp$ is surjective.


