Let a short exact sequence $$ 0 \to L \to M \to N \to 0 $$
is a short exact sequence of $G$-modules, then a long exact sequence is induced:
$$ 0\longrightarrow L^G \longrightarrow M^G \longrightarrow N^G \overset{\delta^0}{\longrightarrow} H^1(G,L) \longrightarrow H^1(G,M) \longrightarrow H^1(G,N) \overset{\delta^1}{\longrightarrow} H^2(G,L)\longrightarrow \cdot $$
The connecting homomorphism is, $$ \delta^n : H^n (G,N) \to H^{n+1}(G, L) $$
Question 1: How to prove the above long exact sequence is true? Is this simply based on the Snake Lemma [It's My Turn (1980)]? Or is there other simpler way to think about it, without using the Snake Lemma?
Let's see that the long exact sequence theorem is equivalent to the snake lemma.
From the short exact sequence we have an induced diagram
$$\require{AMScd}\begin{CD}&&L^n/B^n(G,L)@>>>L^n/B^n(G,M)@>>>L^n/B^n(G,N)@>>>0\\&@VVV@VVV@VVV\\0@>>>Z^{n+1}(G,L)@>{}>>Z^{n+1}(G,M)@>>>Z^{n+1}(G,N)\end{CD}$$
Applying snake lemma we have an exact sequence
$$H^n(G,L)\rightarrow H^n(G,M)\rightarrow H^n(G,N)\xrightarrow{\delta^n}H^{n+1}(G,L)\rightarrow H^{n+1}(G,M)\rightarrow H^{n+1}(G,N).$$
For other hand, a commutative diagram of $G$-modules
$$\require{AMScd}\begin{CD}0@>>>A^0@>>>B^0@>>>C^0@>>>0\\&@V{f}VV@V{g}VV@V{h}VV\\0@>>>A^1@>>>B^1@>>>C^1@>>>0\end{CD}$$
can be viewed as a short exact sequence between complexes $0\rightarrow A\rightarrow B\rightarrow C\rightarrow0$. Then there is an exact sequence between its cohomologies
$$0\rightarrow\ker(f)\rightarrow\ker(g)\rightarrow\ker(h)\xrightarrow{\delta^0}coker(f)\rightarrow coker(g)\rightarrow coker(h)\rightarrow0$$