Lagrangian intersection Floer homology: understanding some assumptions

Let $$(X,\omega)$$ be a symplectic manifold and $$L\subset X$$ be a Lagrangian subspace. Let $$\mu_L:H_2(X,L;\mathbb{Z})\to \mathbb{Z}$$ be the Maslov index homomorphism.

Usual hypothesis

Recall that $$L$$ is said to be monotone, if there exists $$c>0$$ such that the following identity holds for all $$\beta\in \pi_2(X,L)$$: $$c \mu_L(\beta)=\omega(\beta).$$ The minimal Maslov number is defined to be: $$\inf \{\mu_L(\beta) \ |\ \beta\in \pi_2(X,L), \ \omega(\beta)>0 \}.$$ Now given two monotone Lagrangians $$L_0,L_1$$ call assumption $$A1$$ $$(A1): \text{the minimal Maslov number of L_0 and L_1 are strictly greater than 2}$$ and assumption $$A2$$ $$(A2): \text{ L_0 is Hamiltonian isotopic to L_1}\phantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaa}$$

In defining Lagrangian intersection Floer homology groups $$HF(L_0,L_1)$$ usually $$A1$$ or $$A2$$ is assumed.

1. how are these assumption exploited in the construction of the homology? and what's the role of monotonicity?
2. Why assuming the Lagrangians to be spin it is said to simplify things? How can we use the spin assumption?

When you try and prove that $$d^2=0$$ ($$d$$ being the Floer differential) you need to look at the boundary of the moduli space of index 2 J-holomorphic strips with one boundary on $$L_0$$, one on $$L_1$$. Certainly one component of the boundary will consist of "broken strips", i.e. pairs of index 1 strips with a common asymptote that contribute to $$d^2$$. If that were everything, we'd immediately conclude $$d^2=0$$. However, you could bubble off a disc at the side of the strip. The index of the disc and the strip have to add up to 2, but if you have e.g. Maslov 0 discs this could happen without affecting the index of the strip. Monotonicity ensures that zero index discs have zero area, so have to be constant and so don't bubble (they wouldn't eat up any energy/wouldn't be stable).
Maslov 1 discs only occur for nonorientable Lagrangians, so let's ignore those for now. Maslov 2 discs would leave index zero strips, which have to be constant (higher Maslov discs would give negative index strips, which don't exist generically, so don't contribute). This means that $$d^2x$$ ends up being a multiple of $$x$$, the multiple being the count of Maslov 2 discs through $$x$$ which bubble on $$L_0$$ minus the number of discs which bubble on $$L_1$$ (discs can bubble on either side of the strip).
If $$L_0=L_1$$ these numbers are the same, so $$d^2=0$$.
If the minimal Maslov index is bigger than 2 then these numbers are both zero, so again $$d^2=0$$.
In general, you either restrict to Lagrangians with the same number of Maslov 2 discs through each point, or you come to terms with the fact that $$d^2$$ is not zero and your Fukaya category is "curved".