Here is the question and my solution.
I understood the answer discussed here. My question and the solution is slightly different. Which does not use that $T$ is non abelian.
Proof :
CLAIM-1: $T^G$ is direct product simple groups isomorphic to T.
Suppose $T \triangleleft \triangleleft H \triangleleft G$ and let $K=T^H$. Then by induction $K=S_1\times⋯\times S_r$ is minimal normal in $H$, where the $S_i$ are the conjugates of $T$ in $H$.
Now let $K=K_1,K_2,…,K_t$ be the distinct conjugates of $K$ in $G$. These are all minimal normal in $H$, and disjoint. $T^G= K^G = \langle K_1,K_2,...K_t \rangle$.
Claim: $\langle K_1,K_2,...K_t \rangle = K_{j_1} \times K_{j_2} \times ...\times K_{j_m}$ where $j_{1},...j_{m} \in \{1,2,...,t\}$.
Proof: Inductively assume that, $\langle K_2,...K_t \rangle = K_{j_2} \times ...\times K_{j_m}$ for some $j_{2},...j_{m}$.
In the inductive step,
Notice that, $K_1 \cap \langle K_2,...K_t \rangle$ is either $e$ or $K_1$ because $K_1$ is minimal normal subgroup in $H$.
if $K_1 \cap \langle K_2,...K_t \rangle = e$ then $$\langle K_1,K_2,...K_t \rangle = K_{1} \times K_{j_2} \times ...\times K_{j_m}$$
if $K_1 \cap \langle K_2,...K_t \rangle = K_1$, then $K_1 \subset \langle K_2,...K_t \rangle$ and hence $$\langle K_1,K_2,...K_t \rangle = K_{j_2} \times ...\times K_{j_m}$$ Hence CLAIM-1 is proved.
The proof so far does not use that $T$ is non abelian. And hence it is true even when $T$ is abelian. Thus, we can say that $T^G$ is direct product of conjugates of $T$ in G. When $T$ is abelain we get $T^G$ is elementary abelain. Which contradicts part (iii) of the exercise.
Claim 2: $T^G$ is minimal normal when $T$ is non-abelian simple.
The proof of minimality given here works and hence the claim 2 is proved.
Minimality of $T^G$ in G when T is abelian: We do not need to show this, as it wasn't asked in the exercise part (ii).
I am unable to find the mistake in my proof of CLAIM-1. It is incorrect otherwise part(iii) won't be true. Kindly help me with that. Thank you so much
EDIT
Also, can you help to prove part(ii) and part(iii) of the above exercise.
Thanks @DerekHolt for pointing out the mistake in my proof.
For part (ii), suppose that $1 \lhd T = H_1 \lhd H_2 \cdots \lhd T_n = G$.
Then $T \le O_p(H_2)\ {\rm char}\ H_2$, so $O_p(H_2) \unlhd H_3$ and hence $O_p(H_2) \le O_p(H_3)\ {\rm char}\ H_3$, etc, and we end up with $T \le O_p(H_2) \le O_p(H_3) \le \cdots \le O_p(G)$, which is a $p$-group.
For part (iii), let $T$ be a non-normal subgroup of order $2$ in a dihedral group of order $2^k$ with $k \ge 4$.