Direct sum of subspaces projection

Question

Let $V$ be the space of continuous functions $f: [-1, 1] \to \mathbb{R}$. Define $P_E: V \to V$ and $P_O: V \to V$ by $\forall x \in [-1, 1], \forall f \in [-1, 1],$

\begin{align*}\left(P_E f\right)\left(x\right)&=\frac{1}{2}\left[f\left(x\right)+f\left(-x\right)\right] & & & & (1) \\ \left(P_Of\right)\left(x\right)&=\frac{1}{2}\left[f\left(x\right)-f\left(-x\right)\right] & & & & (2) \end{align*}

1. Show that $P_E$ and $P_O$ are linear transformations; then that they are projections; and finally that $P_O = I - P_E$; where $I$ is the identity transformation.
2. How are the nullspaces and the images of $P_E$ and $P_O$ related to each other?
3. Let $P_n$ be the vector space of polynomials with degree less than or equal to n. Compute $P_E(p)$ and $P_O(p)$ for $p \in P_n$. Can you explain why $P_E$ and $P_O$ are called projections onto the even part and the odd part, respectively?

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I did (1) + (2) and (1) - (2), and got $$f\left(x\right)=\left(P_E f\right)\left(x\right)+\left(P_Of\right)\left(x\right),$$ and $$f\left(-x\right)=\left(P_Ef\right)\left(x\right)-\left(P_Of\right)\left(x\right)$$

Not sure how to go from there.