Coin Toss Question with Normal Distribution. Toss a coin and if it lands on heads, then X is distributed normal with Mean=1, Variance=1. If it lands on tails, X is distributed normal with Mean=-1, Variance=1.
(a) Given X=1, what is the chance the coin came up heads?
What I don't understand about this part is that isn't X a continous r.v so P(X=1)=0? How would you condition on such a statement.
(b) What is E[X]?
Thinking intuitively I would guess the answer to this is 0, since the coin is fair and E[X] would be the average of the two means given. Is there a formal approach to finding E[X]?
 A: 

(a) Given X=1, what is the chance the coin came up heads?

What I don't understand about this part is that isn't X a continous r.v so P(X=1)=0? How would you condition on such a statement.

By using conditional density functions. (Provable by taking limits and applying L'Hopital's Rule.) 
$$\mathsf P(H\mid X=x) = \dfrac{f_X(x\mid H)\mathsf P(H)}{f_X(x\mid H)\mathsf P(H)+f_X(x\mid T)\mathsf P(T)}$$
For the given conditional distributions, we have $f_X(x\mid H) = e^{-(x-1)^2/2}/\sqrt{2\pi}$ and $f_X(x\mid T) = e^{-(x+1)^2/2}/\sqrt{2\pi}$.
If the coin is unbiased, $\mathsf P(H)=\mathsf P(T)$ so then $$\begin{align}
\mathsf P(H\mid X=x)
 & = \dfrac{f_X(x\mid H)}{f_X(x\mid H)+f_X(x\mid T)}
\\[1ex]
 & = \dfrac{e^{-(x-1)^2/2}}{e^{-(x-1)^2/2}+e^{-(x+1)^2/2}}\color{silver}{\require{cancel}\cdot\frac{\bcancel{1/\sqrt{2\pi}}}{\bcancel{1/\sqrt{2\pi}}}}
\\[2ex]
\mathsf P(H\mid X=1) & = \dfrac{1}{1+e^{-2}}
\end{align}$$


(b) What is E[X]?

Thinking intuitively I would guess the answer to this is 0, since the coin is fair and E[X] would be the average of the two means given. Is there a formal approach to finding E[X]?

That is sufficient.   More formally, your intuitive approach is an application of the Law of Total Expectation. (a.k.a the Law of Iterated Expectation).
$$\begin{align}
\mathsf E(X) & = \mathsf E(X\mid H)\mathsf P(H)+\mathsf E(X\mid T)\mathsf P(T)
\\ & = 1\times \tfrac 1 2 + (-1)\times \tfrac 1 2
\\ & = 0
\end{align}$$
