If $M \sim \cal N(0,1)$ and $X$ conditioned on $M$ is normally distributed with expected value $\alpha M$, $\alpha>0$ and variance $\sigma^2>0$, how do I know if $X$ is normally distributed?
Also is this right:
$E(X)=E(E(X \vert M)) = E(\alpha M)=\alpha E(M)= \alpha \times 0=0 $ and $V(X)=E(X^2)-E(X)^2=E(E(X^2\vert M^2))=E(\alpha^2M^2)=\alpha^2 E(M^2)=\alpha \times 1=\alpha$?
Your calculation of the variance has a couple of errors. The simple one is when you say $\alpha^2 E(M^2)=\alpha \times 1$ instead of $\alpha^2 E(M^2)=\alpha^2 \times 1$. The slightly more complicated one is where you say $E(E(X^2\mid M^2))=E(\alpha^2M^2)$ instead of $E(E(X^2\mid M)) = E(\sigma^2 + \alpha^2M^2)$. Correct these and you get $\sigma^2+\alpha^2$ as the variance of $X$
Another possible approach:
$\alpha M$ has a normal distribution with expected value $0$ and variance $\alpha^2$
$Y=X-\alpha M$ has a normal distribution with expected value $0$ and variance $\sigma^2$, and is independent of $\alpha M$
so $X=Y+\alpha M$ has a normal distribution with expected value $0+0=0$ and variance $\sigma^2+\alpha^2$