Let $P_1 $ and $ P_2 $ be linear operators in a vector space $ E $ such that:
$$P_1+P_2=I,\quad P_1P_2=0\quad P_2P_1=0,\quad P_1^2=P_1,\quad \mbox{and}\quad P_2^2=P_2;$$ Show that $Im(P_1)=N(P_2)$.
I have doubts about how to use the hypotheses, I don't know how to construct the subspaces $ F_1 $ and $ F_2 $ such that $ E $ is a direct sum of them. To later conclude that $ P_1 $ and $ P_2 $ are projections and thus be able to make the final demonstration.
If $x \in E$, then $P_2P_1x = 0$ since $P_2P_1 = 0$. Hence, $\operatorname{Im}(P_1)\subseteq N(P_2).$
Conversely, assume that $P_2(x) = 0$ for some $x \in E$. Then $$x= Ix = (P_1 + P_2)(x) = P_1(x) + P_2(x) = P_1(x)$$ and thus $N(P_2) \subseteq \operatorname{Im}(P_1).$
Combining these two inclusions, we obtain $\operatorname{Im}(P_1) = N(P_2).$
On a side note, you can drop some assumptions. I only used the assumptions $P_2P_1 = 0$ and $P_1 + P_2 = I.$