Suppose first that $M\perp N$. Then, for any $x,y\in H$,
$$
\langle P_MP_Nx,y\rangle=\langle P_Nx,P_My\rangle=0.
$$
As we can do this for all $x,y$, we get that $P_MP_N=0$. Then
$$
(P_M+P_N)^2=P_M^2+P_N^2=P_M+P_N
$$
and $P_N+P_M$ is a projection. Given $x+y\in M+N$ with $x\in M$ and $y\in N$, $$(P_M+P_N)(x+y)=P_Mx+P_Nx+P_My+P_Ny=x+y,$$since $P_Mx=x$, $P_nx=0$, $P_My=0$, $P_Ny=y$. If $z\in (M+N)^\perp$, then $$\langle(P_M+P_N)z,w\rangle=\langle z,(P_M+P_N)w\rangle=0.$$ So $P_M+P_N=P_{M+N}$.
Conversely, if $P_M+P_N=P_{M+N}$, then
$$
P_M+P_N=P_{M+N}=P_{M+N}^2=P_M+P_N+P_NP_M+P_MP_N.
$$
Then $$\tag{*}P_MP_N+P_NP_M=0.$$ Multiply on the left by $I-P_M$, and we get $(I-P_M)P_NP_M=0$. This is $P_MP_NP_M=P_NP_M$. Then
$$
P_MP_N=(P_NP_M)^*=(P_MP_NP_M)^*=P_MP_NP_M=P_NP_M.
$$
Then $P_NP_M=P_MP_N$ and now $(*)$ becomes $P_MP_N=0$. If we take $x\in M$ and $y\in N$,
$$
\langle x,y\rangle=\langle P_Mx,P_Ny\rangle=\langle P_NP_Mx,y\rangle=0.
$$
So $M\perp N$.
Let me prove (i) $\Rightarrow$ (ii), while giving you some more time to think about (iii) $\Rightarrow$ (i), which is
not so hard.
By hypothesis $$
P_1+P_2=
(P_1+P_2)^2 =
P_1^2+P_1P_2+P_2P_1+P_2^2 =
P_1+P_1P_2+P_2P_1+P_2.
$$
so $P_2P_1=-P_1P_2$. Left multiplying the above by $P_2$ we get
$$
P_2P_1=-P_2P_1P_2.
$$
Since $P_2P_1P_2$ is self-adjoint, then so is $P_2P_1$. But then
$$
P_1P_2= P_1^*P_2^*= (P_2P_1)^* = P_2P_1 = -P_1P_2,
$$
whence $P_1P_2=0$.
Best Answer
If $PQ$ is an orthogonal projection then in particular $PQ = (PQ)^\ast$, hence $$PQ = (PQ)^\ast = Q^\ast P^\ast = QP$$ because $P = P^2 = P^\ast$ and $Q = Q^2 = Q^\ast$ by hypothesis. Thus $M \ni PQx = QPx \in N$ and hence $PQ(\mathcal{H}) \subset M \cap N$. For all $x \in M \cap N$ we have $x = Px$ and $x = Qx$, hence also $x = PQx$, so $PQ(\mathcal{H}) = M \cap N$. This shows the first claim as well as one direction of the second assertion.
For the other direction verify that $PQ = (PQ)^2 = (PQ)^\ast$.