Converting S->aTbS|epsilon T->aTb|epsilon to chomsky normal form
Our grammar $G$ has following two production where $S$ is start symbol.

$S\rightarrow aTbS \vert\epsilon$

$T\rightarrow aTb|\epsilon$

Now to convert it to chomsky normal form (CNF) we have to perform following steps:


If there is some production having $S$ in it's right side($S\rightarrow aTbS$ in this example) then add new start symbol $S_0$ and production rule $S_0 \rightarrow S$ to $G$.
Then remove null production from $G$:
So, after removing null production $G$ will have following productions:

$S_0 \rightarrow S$

$S \rightarrow aTbS|abS|aTb|ab$

$T \rightarrow aTb|ab$
Now we have to remove unit production from $G$. But there are none of them.
Then if there is production containing both terminal and nonterminals on RHS then introduce new non-terminal for the terminal appearing there and replace occurence of terminal in that RHS by this non-terminal. example will make this clear.

we, have this $S \rightarrow aTbS$ production. So we add production $A \rightarrow a$
and $B\rightarrow b$ to G. and then rewrite production $S \rightarrow aTbS$ as $S \rightarrow ATBS$

Doing same for all production we will have production set of $G$ as follow:

$S_0 \rightarrow S$

$S \rightarrow ATBS|ABS|ATB|AB$

$T \rightarrow ATB|AB$

$A \rightarrow a$

$B \rightarrow b$
Now we handle the case where RHS of production contains more than 2 non-terminals.
What we do is simple. We just keep first non-terminal as it is and introduce new 
nonterminal for remaining part. Again example will make this clear.

We have this $S \rightarrow ATBS$ production. So we introduce new non-terminal $C$ 
and add production $C\rightarrow TBS$. Then modify original production as 
$S\rightarrow AC$.

Note here that now we have to do this process for newly introduced production also. 
i.e $C\rightarrow TBS$ will be converted to $C\rightarrow TD$ for instance and we 
have to add $D \rightarrow BS$

Now after this step $G$ will contain following productions:

$S_0 \rightarrow S$

$S \rightarrow AC|AD|AE|AB$

$C \rightarrow TD$

$D \rightarrow BS$

$E \rightarrow TB$

$T \rightarrow AE|AB$

$A \rightarrow a$

$B \rightarrow b$


So now grammar is in CNF. But one thing is still remaining. Original grammar produces empty string but this grammar does not. For that add production $S_0 \rightarrow \epsilon$ to $G$.