I will add some detail to the previous answers by abx, Robert Bryant (in the linked question), and V. S. Matveev.
Regarging question 1: A simply connected manifold $M$, admits a flat connection if an only if it is paralellizable.
Indeed, you can parallel-transport a frame (a basis of the tangent space) from some initial point $p$ to every point $q$ along any curve $\gamma$. The fact that the connection is flat implies that the frame that you obtain at the endpoint $q$ won't vary if you perform a continuous variation (homotopy) of the curve $\gamma$; it only depends on the homotopy class. But there's only one homotopy class, so the frame at $q$ is independent of the choice of $\gamma$, and you get a basis of vector fields $X_i$ that are parallel with respect to the connection: $\nabla_YX_i=0$ for all $X_i$ and $Y$. For examples, observe that all the spheres $S^n$ with $n\geq 2$ are simply connected, and the only parallelizable spheres are $S^1$, $S^3$ and $S^7$. If $n$ is even, you can quickly see that there is not even one nonvanishing vector field, by the Poincaré-Hopf theorem, because the Euler characteristic is 2.
Regarding question 3: You can furthermore prove that spheres of dimenison $n\geq 2$ don't admite _symmetric_ flat connections.
Indeed, if the connection was symmetric (zero torsion), then the vector fields that you obtain in last construction commute: $[X_i,X_j]=0$. Now, since each $X_i$ gives rise to an action of $\mathbb R$ on the manifold $M$, and the actions commute, you get an action of $\mathbb R^n$ by "translations" in the directions of the vector fields $X_i$, the translation by a vector $w\in\mathbb R^n$ being the result of travelling along $M$ in the direction of the vector field $\sum_i w_iX_i$ during one unit of time. The orbits of this action are open sets because the action is free (you can "wiggle" a point in every direction because the vector fields form a basis of the tangent space), and there is only one such orbit because $M$ has one connected component (so the action is transitive). The only compact $n$-manifold that admits a free, transitive Lie group actions of $\mathbb R^n$ is the $n$-torus, which is not simply connected. So a compact, simply connected manifold can't have a symmetric flat connection.
Regarding question 2: To see that $S^2\times S^1$ does not admit a flat metric, observe that if it did, then by the Cartan-Hadamard theorem, its universal cover would be diffeomorphic to $\mathbb R^3$. But it's universal cover is $S^2\times\mathbb R$.
As V.S. Matveev observes in his comment below, the Cartan-Hadamard argument can also be used to prove that spheres don't admit a symmetric flat connection.