Is projection of a measurable subset in product $\sigma$-algebra onto a component space measurable?

The answer to the second question is "no". Take $\Omega_1=\Omega_2=\{0,1\}$ and let ${\cal F}_1=\{\emptyset,\{0\},\{1\},\{0,1\}\}$ and ${\cal F}_2=\{\emptyset,\{0,1\}\}$. The diagonal in $\Omega_1\times\Omega_2$ is not measurable with respect to the product $\sigma$-algebra ${\cal F}_1\times {\cal F}_2$, but its projection either way is the whole space.

The answer to Ethan's Question 1 is no. I n descriptive set theory, a subset $A$ of a Polish space $X$ is an analytic set if it is a continuous image of a Polish space. These sets were first defined by Luzin (1917) and his student Souslin (1917)(see, http://en.wikipedia.org/wiki/Analytic_set ).

We need well known facts from the descriptive set theory.

Fact 1. The following conditions on a subset $A$ of a Polish space $X$ are equivalent:

(a) $A$ is analytic;

(b) There is a Polish space $Y$ and a Borel set $B \subseteq X \times Y$ such that $A$ is projection of $B$., that is $A=\{ x \in X | (\exists y)(x,y) \in B\}$.

Fact 2. There exists an analytic set $A_0$ in a Polish space $X$ which is not Borel.

Let $A_0$ becomes from Fact 2. For $A_0$, the condition (b) of Fact 1 implies that there exists a Polish space $Y_0$ and a Borel set $B_0 \subset X \times Y_0 $ such that $A_0=\{ x \in X | (\exists y)(x,y) \in B\}$. Obviously, $B_0$ stands an example of Borel subset of $X \times Y_0$ whose projection on $X$ is a non-Borel set $A_0$.
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The answer to Ethan's Question 2 is no. Let $Y$ be a non-Borel subset of $[0.1]$.

Let consider a set

$C:=([0,1]\times [0,1]) \setminus ((\{1/2\}\times Y) \cup (Y \times \{1/2\}))$.

Then its projection onto any component space defined as above is Borel measurable(more precisely, coincides with corresponding component space) but $C$ is not Borel measurable.