You are being careless with equalities and isomorphisms.
Let $G=S_1\times\cdots \times S_n$ where each $S_i$ is simple. Let $\iota_k\colon S_k\hookrightarrow G$ be the canonical embeddings, sending $g\in S_k$ to the tuple $(e,\ldots,g,\ldots, e)$ that has $g$ in the $k$th coordinate.
Your "Claim 1" should be:
Claim 1: For each $k$, $\iota_k(S_k)$ is a minimal normal subgroup of $G$.
This is true because each $\iota_k(S_k)$ is normal, and any normal subgroup of $G$ contained in $\iota_k(S_k)$ corresponds to a normal subgroup of $\iota_k(S_k)\cong S_k$, hence is trivial or all of $S_k$.
Claim 2: If each $S_k$ is nonabelian, then the only normal subgroups of $G$ are of the form $\iota_{i_1}(S_{i_1})\cdots\iota_{i_r}(S_{i_r})$ for some $1\leq i_1\lt\cdots\lt i_r\leq n$, $0\leq r\leq n$.
These are all normal, being products of normal subgroups. Now assume that $N\triangleleft G$. Then $\pi_j(N)\triangleleft S_j$, hence the projection is either trivial or all of $S_j$. Thus, we may assume that $N$ is a subdirect product $G=S_1\times\cdots\times S_n$, and we aim to show that it is then equal to $G$.
I claim that $\iota_1(S_1)\subseteq N$. Indeed, let $x,y\in S_1$. Because $N$ is a subdirect product, there exists an element in $N$ with first component equal to $x$, say $(x,s_2,\ldots,s_n)\in N$. Conjugating with $\iota_1(y)=(y,e,e,\ldots,e)$, we have that $(yxy^{-1},s_2,\ldots,s_n)\in N$. Now multiplying by $(x,s_2,\ldots,s_n)^{-1}\in N$ on the right we obtain
$$(yxy^{-1}x^{-1},e,\ldots,e)= ([y,x],e,\ldots,e)\in N.$$
Thus, $\iota_1([S_1,S_1])\subseteq N$. But because $S_1$ is nonabelian, $S_1=[S_1,S_1]$; thus, $\iota_1(S_1)\subseteq N$.
Similarly, $\iota_k(S_k)\subseteq N$ for each $k$; thus, a generating set for $G$ is contained in $N$, so $N=G$, as desired.
Corollary. If all $S_k$ are nonabelian and simple, then the minimal normal subgroups of $S_1\times\cdots \times S_n$ are precisely the subgroups $\iota_k(S_k)$, for $k=1,\ldots,n$.
Claim 2 is false if you have two isomorphic abelian factors, as then you have "diagonal" subgroups, like $\langle (x,x)\rangle\leq C_2\times C_2$.
As stated your Claim 1 is false, because you may have subgroups isomorphic to some of the $S_i$ which are not normal. For example, if there is any $S_k$ that is isomorphic to (a subgroup of) $S_j$ for some $k\neq j$ via a morphism $\phi$, then the subgroup generated by the elements $\iota_k(x)\iota_j(\phi(x))$ is a subgroup that is isomorphic to $S_k$, but will not be normal in $G$.
Let's look a bit more carefully at what happens when you have abelian factors. Suppose that
$$G=S_1\times\cdots\times S_n\times A_{n+1}\times\cdots\times A_{n+m}$$
where the $S_i$ are simple and nonabelian, and $A_{n+1},\ldots,A_n$ are simple abelian (hence cyclic of prime order).
Lemma. If $N\triangleleft G$ and there exists $k$, $1\leq k\leq n$ such that $\pi_k(N)\neq\{e\}$, then $\iota_k(S_k)\leq N$.
Proof. The argument is similar to the one for Claim 2 above: given $x,y\in S_k$, we have an element $n\in N$ with $\pi_k(n)=x$. Then $\iota(y)n\iota(y)^{-1}n^{-1}$ has $k$th entry equal to $[y,x]$, and all other entries trivial, so $\iota_k([S_k,S_k])\leq N$; but this equals $\iota_k(S_k)$ because $S_k$ is simple and nonabelian. $\Box$
Corollary. If $N\triangleleft G$, and $(s_1,\ldots,s_n,a_1,\ldots,a_m)\in N$, then $(s_1,\ldots,s_n,e,\ldots,e)\in N$ and $(e,\ldots,e,a_1,\ldots,a_m)\in N$.
Proof. If any $s_j\neq e$, then $N$ contains $\iota_j(S_j)$, Thus, it contains $(s_1,\ldots,s_n,e,\ldots,e)$. $\Box$
Corollary. If $N\triangleleft G$, then $N=M\times B$, where $M\triangleleft S_1\times\cdots\times S_n$, and $B\triangleleft A_{n+1}\times\cdots\times A_{n+m}$.
Now, we already know what the normal subgroups of $S_1\times \cdots\times S_n$ are, so we know what $M$ looks like.
For $B$, we have a finite abelian group, which is isomorphic to $E_{p_1,k_1}\times\cdots\times E_{p_r,k_r}$, where $p_1,\ldots,p_r$ are pairwise distinct primes, $k_i\geq 1$, and $E_{p_i,k_i}$ is an elementary abelian $p_i$-group of rank $k_i$; that is, $E_{p_i,k_i}\cong (C_{p_i})^{k_i}$. We know that $B$ is the product of its $p_i$ parts, so we are left with considering the subgroups of $E_{p_i,k_i}$. From linear algebra we know they just elementary abelian, isomorphic to $(C_{p_i})^{t_i}$ with $0\leq t_i\leq k_i$.
In particular, any minimal normal subgroup will be isomorphic to $C_{p_i}$, hence to one of the factors.
So we have:
Theorem. Let $G= S_1\times\cdots \times S_n$, where each $S_i$ is a finite simple group. Then the minimal normal subgroups of $G$ are precisely the subgroups $\iota_k(S_k)$ if and only if for each prime $p$ there is at most one $S_i$ that is cyclic of order $p$. If there are other minimal normal subgroups of $G$, then they are isomorphic to a cyclic group of order $p$, and so are at least two of the $S_k$.
Best Answer
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 $k \ge 4$.