$1.$ Lichnerowicz proved that if $M$ is a closed spin manifold which admits a positive scalar curvature metric, then $\hat{A}(M) = 0$. In dimensions $4k$, $\alpha(M) = 2\hat{A}(M)$, so it follows that $\alpha(M) = 0$. In fact, Hitchin proved that $\alpha(M) = 0$, regardless of dimension.
$2.$ In dimension four, $\hat{A}(M) = -\frac{1}{24}p_1(M) = -\frac{1}{8}\tau(M)$, so $\alpha(M) = 0$ if and only if $\tau(M) = 0$. If $M$ is a simply connected, closed, spin four-manifold with signature zero, then by Freedman's classification, there is an integer $k \geq 0$ such that $M$ is homeomorphic to $k(S^2\times S^2)$, the connected sum of $k$ copies of $S^2\times S^2$; note, the connected sum of zero copies of $S^2\times S^2$ is defined to be $S^4$. If $M$ is diffeomorphic to $k(S^2\times S^2)$, then $M$ admits positive scalar curvature metrics. Said another way, the topological manifolds $k(S^2\times S^2)$ all admit positive scalar curvature for their standard smooth structure.
A closed four-manifold can potentially admit a countably infinite number of smooth structures, and the existence of a positive scalar curvature metric depends on which smooth structure you choose. So it is possible that for some choice of $k$ and some choice of smooth structure on $k(S^2\times S^2)$, there does not admit a positive scalar curvature metric. Such examples exist. In Hirzebruch Surfaces: Degenerations, Related Braid Monodromy, Galois Covers (MR 1720873), Teicher constructs examples of simply connected general-type complex surfaces which are spin and have signature zero; see Theorem $5.8$. These are homeomorphic to $k(S^2\times S^2)$ for some $k$, but general type surfaces do not admit positive scalar curvature metrics.
$3.$ In dimension four, unlike in other dimensions, there are gauge theoretic obstructions to the existence of positive scalar curvature metrics. In particular, on a closed smooth four-manifold with $b^+ \geq 2$, the existence of a positive scalar curvature metric implies that all of its Seiberg-Witten invariants vanish. This is a necessary condition, but it is not sufficient. Note that these invariants are usually too complicated to calculate, except in ideal situations such as when the four-manifold in question is the underlying four-manifold of a Kähler surface.
