Let $G$ be a profinite group, then there is a general notion of continuous cohomology groups $H^n_{\text{cont}}(G, M)$ for any topological $G$-module $M$ (I require topological $G$-modules to be abelian). Although this definition makes perfect sense, it has a drawback that it is not a "delta-functor" (the category of topological $G$-modules is not abelian, so I will explain below what I really mean by this). It is stated in various places that this is not true that an arbitrary short exact sequence of topological $G$-modules $$ 0 \to M' \to M \to M''\to 0 $$ induces a long exact sequence of continuous cohomology groups $$ 0 \to H^0_{\text{cont}}(G, M')\to H^0_{\text{cont}}(G, M) \to H^0_{\text{cont}}(G, M'') \to H^1_{\text{cont}}(G, M') \to H^1_{\text{cont}}(G,M) \to \dots $$ However, I've never seen an example of such a short exact sequence for a profinite group $G$. Before asking a precise question, let me explicitly say what I mean by an exact sequence of topological $G$-modules.
Definition: A short sequence of topological $G$-modules $0 \to M' \to M\to M'' \to 0$ is called exact, if it is exact as a sequence of abstract groups, topology on $M'$ is equal to the topology induced from $M$ and the map $M \to M''$ is a topological quotient map.
Question 1: What is an explicit example of a profinite group $G$ and a short exact sequence of topological $G$-modules $0 \to M' \to M \to M'' \to 0$ s.t. there is no associated long exact sequence of continuous cohomology groups?
Question 2: Is there an example of such a sequence with $M'$ being (quasi-)compact?
Apparently, the answer to the second question should positive. And I am more interested in this question, but I don't even know an example for the first question.
Remark: It is easy to show that if $M \to M''$ has a continuous section (as topological spaces), then the associated long exact sequence is exact. So, whatever these examples are the map $M \to M''$ shouldn't have a continuous section. In particular, they can't be finite-dimensional spaces over a field and $M''$ can't be a discrete module. Also, if $M'$ and $M$ itself are profinite groups, then Lemma 5.6.5 from the book "Profinite groups" by Ribes and Zalesskii guarantees that $M \to M''$ has a continuous section.
Remark 2: If you drop the assumption that $G$ is a profinite group, then it is easy to construct an example. Namely, consider $G=S^1$ and a short exact sequence $$ 0 \to \mathbf Z \to \mathbf R \to S^1 \to 0 $$ with trivial $S^1$ action. It is not hard to show that $H^1_{\text{cont}}(S^1, \mathbf R)=H^2_{\text{cont}}(S^1, \mathbf Z)=0$ but $H^1_{\text{cont}}(S^1, S^1)=\operatorname{Hom}_{\text{cont}}(S^1,S^1)\neq 0$.
