curvature and volume growth 
*

*Let $M$ be a non-compact connected Riemannian manifold with $\mathrm{sec}_g=0$ and $\operatorname{vol} B(x,r)\geq c(n)r^n$ for any $r$, where $c(n)>0$. How to prove that $(M,g)$ is isometric to $(R^n,\mathrm{can})$?

*Let $M$ be a non-compact connected Riemannian manifold with $\mathrm{ric}_g=0$, $\mathrm{conj.rad}=\infty$, and $\operatorname{vol}B(x,r)\geq c(n)r^n$ for any $r$, where $c(n)>0$. How to prove that $(M,g)$ is isometric to $(R^n,\mathrm{can})$?
 A: Answer to the first question :
a complete flat Riemannian manifold of dimension $n$ is a quotient $\mathbb{R}^n$ by a discrete subgroup of isometry $\Gamma$.
The Euclidean volume growth implies that $\Gamma$ is finite but as $\Gamma$ acts freely on $\mathbb{R}^n$, we must have $\Gamma=\{1\}$.
About the second question :
According to an inequality of Calabi  and Yau, the volume growth of a complete Riemannian manifold of non negative Ricci curvature is contoled by
$$\forall r\ge 1 :{\mathrm vol}B(x,r)\ge k(n)r {\mathrm vol} B(x,1).$$
According to M. Anderson and P. Li a complete Riemannian manifold with non negative curvature and Euclidean growth is simply connected.
(M. Anderson, On the topology of complete manifolds of nonnegative Ricci curvature, Topology 29 (1990) 41--55 and  P. Li, Large time behavior of the heat equation on complete manifolds with nonnegative Ricci curvature, Ann. of Math. 124 (1986) 1-21.)
In your case, the manifold is necessary diffeomorphic to $\mathbb{R}^n$. Because the hypothesis on the conjugate radius implies that the exponential map is a covering.  As the Euclidean volume growth implies that the fundamental group is finite, this will imply that 
this covering is trivial.
There are metric on $\mathbb{R}^n$ that are rotationnally symmetric, with non negative ricci curvature and Euclidean growth for instance in dimension $2$ take the metric (with $a\in (0,1)$)
$$g=(dr)^2+\left(a r+(1-a)\tanh(r) \right)^2(d\theta)^2$$
The conjugate radius at the  $0$ is infinite.
If you assume that the  conjugate radius is infinite at every point of the manifold. This implies that all geodesic must be a ray :
let $\gamma\colon \mathbb{R}\mapsto M$ be a infinite geodesic.
$$\gamma(t)=\exp_{\gamma(0)} (tv)$$
We know that $\gamma$ is minimizing on $[0,+\infty)$ and on $(-\infty,0]$ because the exponential map is a diffeomorphism.
But we also have for all $T$ :
$$\gamma(t)=\exp_{\gamma(T)} ((t-T)\dot \gamma(T))$$
Hence $\gamma$ is minimizing on $[T,+\infty)$ and on $(-\infty,T]$ hence $\gamma$ is a ray and the Cheeger-Gromoll splitting theorem implies that the manifold is isometric to the Euclidean space.
