Let $BG$ is classifying space of $G$ topological group.


If $G$ is any compact group and $H$ is a closed subgroup of $G$, then the
inclusion map $i:H\rightarrow G$ induces 
\begin{equation\*}
G/H\rightarrow BH\rightarrow BG
\end{equation\*}
a fiber bundle?


If $G$ is any compact group and $H$ is a closed subgroup of $G$, then the
inclusion map $i:H\rightarrow G$ induces  

\begin{equation\*}
G/H\rightarrow BH\rightarrow BG
\end{equation\*}
a fibration?


If $G$ is any compact group and $N$ is a closed normal subgroup of $G$, then
the quotient map $\pi :G\rightarrow G/N$ induces  

\begin{equation\*}
BN\rightarrow BG\rightarrow B\left( G/N\right) 
\end{equation\*}
a fiber bundle?


If $G$ is any compact group and $H$ is a closed normal subgroup of $G$, then
the quotient map $\pi :G\rightarrow G/N$ induces 
\begin{equation\*}
BN\rightarrow BG\rightarrow B\left( G/N\right) 
\end{equation\*}
a fibration?



Fiber bundle and fibration of classifying space
No, this is not true for Q1 and Q2.


To be specific we probably need to pick a particular model for $BG$, and there are several. For the argument I will give, the following property is sufficient: there exist isomorphisms $\pi\_{i+1}(BG) \cong \pi\_i(G)$ of homotopy groups.


Let $G = S^1$ and let $H$ be the subgroup of torsion elements (the roots of unity). The group $G$ is abelian so $H$ is normal, and $H$ is totally disconnected so $\pi\_0(H) \cong H$ as sets. Any open set in $G$ intersects any coset $gH$ nontrivially, so the quotient space $G/H$ has the indiscrete topology and its homotopy groups vanish.


If the answer to any of the first two questions that you asked was affirmative, there would be an exact sequence in homotopy groups of the form
$$
\cdots \to \pi\_1(G/H) \to \pi\_1(BH) \to \pi\_1(BG) \to \cdots
$$
and this would be an exact sequence
$$
\cdots \to 0 \to H \to 0 \to \cdots
$$
which is not possible.


