Finding the conjugate operator of the following operators The operators are : $$
\\\ \\ \\   \ \ 1)\ \ A:\ell_2 \to \ell_2 \  \ \ \ \ \ \ A(x_1,x_2,.,x_n,..)= (\lambda_1 x_1, \lambda_2 x_2,.,\lambda_i x_n,..) \ \ \ ; \ \ \sup|\lambda_i|<\infty
$$
$$\\\ \\ \ \  2)\ \ A:c_0 \to c_0 \ \ \text{same operator}$$
Find their conjugates.

Definition: Let $A: X \to Y$ be an operator where $X,Y$ are Banach spaces.
  The conjugate operator of $A$ is the operator $A^{*}:Y^* \to X^*$ such that: 
  $$A^{*}y^{*}(x)=y^{*}(Ax) \ \ or \\ \langle A^{*}y^{*},x\rangle =\langle y^{*},Ax\rangle $$

I believe this is:
https://en.wikipedia.org/wiki/Hermitian_adjoint
https://en.wikipedia.org/wiki/Riesz_representation_theorem
I have the solution but this does not make any sense to me and I would like it if someone could explain the difference in the solutions, looking at these differing spaces in between which  the operators act upon.
First off we know that $(l_2)^*\cong l_2$ and that $(c_0)^* \cong l_1$.
1.) Solution:
$$ A^*:(l_2)^* \to (l_2)^* \iff A^*:l_2 \to l_2  \\
   A^*y^{*} \in X^* \implies \exists m\in X: m= \{ \eta_i \}_{i=1}^{\infty} \  \  \text{such that}\ \forall x: \ \   A^*y^{*}(x) = \sum_{i=1}^{\infty}\eta_i x_i. \\
  y^*(Ax) = y^*((\lambda_i x_i)_{i=1}^{\infty})=\sum_{i=1}^{\infty}y_i \lambda_i x_i \\ \implies \eta_i = \lambda_i y_i \implies A^{*}y^{*}(x)= \sum_{i=1}^{\infty}y_i \lambda_i x_i  \\ A^{*}y^{*} = ((\lambda_i y_i)_{i=1}^{\infty})^{*} $$
what's with this star above this series?? How is this deducted? Everything is clear up to this point here.
$$A^{*}(y)=(\lambda_1 y_1, \lambda_2 y_2,.., \lambda_n y_n,..) \implies A=A^{*}$$
2.) Solution:(this one makes less sense to me) 
$$A:c_0 \to c_0 \implies A^{*} : l_1 \to l_1 \\ 
  A^*y^*(x) = <A^*y, x>= \sup_{i\in \mathbb{N}}|\eta_i x_i| \\ $$
This is where I am lost alre\dy. On the side it says (in comment) that this is written "conditionally". I do not know what this means, nor do I understand how this is. I know that the norm in $c_0$ is $\sup$, but how that plays a role here is beyond me right now. The solution goes on to:
$$y^*(Ax)=<y,Ax>=\sup_{i\in \mathbb{N}}|\lambda_i y_i x_i|$$
comment: this has to imply for every $x \in l_1$ 
$$\implies |\eta_i|=|\lambda_i y_i|$$
comment: this applies to $e_i=(0,...1,....0)$(i-th coordinate is $1$, the rest are zero) aswell, so when we use $|\eta_i|=sgn y_i y_i |\lambda_i|$ we see that $$A^*y = (\lambda_1 y_1, \lambda_2 y_2,..., \lambda_n y_n,..)$$

I would very much appreciate help on understanding this. This type of assignment was half of the final exam one year, and we only did this assignment in class. Very thankful in advance. 

 A: In the solution to $1$, the star could mean that one sees the dual as the "transpose". It is weird also because $A$ will be selfadjoint only if the $\lambda_j$ are real, which is not stated. 
Here is the way to write it by taking advantage of the fact that $\ell_2^*=\ell_2$.
$$
\langle A^*y,x\rangle = \langle y,Ax\rangle =\overline{\langle Ax,y\rangle}
=\overline{\sum_j (Ax)_j\overline{y_j}}=\overline{\sum_j \lambda_j x_j\overline{y_j}}
=\sum_j\overline{\lambda _j}y_j\,\overline{x_j}=\langle By,x\rangle,
$$
where $$B(x_1,x_2,\ldots)=(\overline{\lambda_1}x_1,\overline{\lambda_2}x_2,\ldots). 
$$
In particular, when the $\lambda_j$ are real, $A^*=A$. 
What I see in 2 doesn't make a lot of sense to me. As mentioned in the comments, the $\sup$ is definitely out of place. Here, for $y\in \ell_1$, and $x\in c_0$, 
$$
A^*y(x) =y(Ax)=\sum_jy_j(Ax)_j=\sum_j y_j\lambda_jx_j=\sum_j(\lambda_jy_j)x_j. 
$$
So $A^*y$, as an element of $\ell_1$, is $$A^*y=(\lambda_1y_1,\lambda_2y_2,\ldots).$$
Here we cannot write $A^*=A$ because they operate on different spaces but we do have, since $\ell_1\subset c_0$, that $A|_{\ell_1}=A^*$. 
