1. Show that if $a: G \times M \to M$ is proper, then $\theta= (\pi_M, a): G \times M \to M \times M $ is proper. But converse is not true.
Take a compact set $K \In M \times M$, then $\theta^{-1}(K) \in a^{-1}( \pi_2(K))$, where $\pi_2: M \times M \to M$ is the projection to the second factor. Since $\pi_2$ is continuous, $\pi_2(K)$ is compact; since $a$ is proper, $a^{-1}(\pi_2(K))$ is compact. $\theta^{-1}(K)$ is closed subset of a compact set, hence is compact.
For a counter-example of the converse statement, consider $G \times G \to G \times G$, where $G = \R$ and action is addition.