Along the lines of this article:
Let $I: H\rightarrow H^*$ map each ket $|\phi\rangle$ to its corresponding bra $\langle \phi |$ and $H$ is a Hilbert space and $H^*$ its dual.
$A: H\rightarrow H$ is an operator on $H$.
$A^\dagger$ (the adjoint of $A$) can be defined as: $|\phi\rangle \rightarrow I^{-1}(\langle\phi|A)$. In other words, $A^\dagger$ maps each ket to the ket corresponding to $\langle \phi |A$. I have two questions:
1 - How can I show that $I(A|\phi\rangle)=\langle\phi|A^\dagger$
2 - How can I show that in general, $I^{-1}(\langle\phi|A)\ne A|\phi\rangle$ or equivalently, $\langle\phi|A\ne I(A|\phi\rangle)$? (my understanding is that in the particular case when equality holds, we call $A$ a Hermitian operator)
Please also let me know if any statements above are incorrect.
Because the inner product in a complex Hilbert space is conjugate symmetric one has the identity $ \langle x|y\rangle=\overline{\langle y|x\rangle}=\overline{I(|y\rangle)|x\rangle} $ and equivalently $\overline{\langle x|y\rangle}=I(|y\rangle)|x\rangle)$.
Applied to our problem this yields $$ \langle \phi|A^\dagger|\psi\rangle=\overline{I(A^\dagger|\psi\rangle)|\phi\rangle} \overset{\text{Def. of }A^\dagger}=\overline{I(I^{-1}(\langle\psi|A))|\phi\rangle} =\overline{\langle\psi|A|\phi\rangle}=I(A|\phi\rangle)|\psi\rangle $$ for all $\phi,\psi$. But because $|\psi\rangle$ was chosen arbitrarily the corresponding elements from $H^*$ have to coincide, i.e. $\langle \phi|A^\dagger=I(A|\phi\rangle)$ so $(A^\dagger)^\dagger=A$ as desired. This argument works analogously for real Hilbert spaces simply by omitting all the complex conjugates.
As for your second question, for a generic operator $A$ we may assume the existence of $\phi,\psi$ such that $\langle \phi|A|\psi\rangle\neq \overline{\langle\psi|A|\phi\rangle}$ (if you want a specific counter-example to see that there is no reason for "a general operator" to satisfy this, choose $A=\scriptsize\begin{pmatrix}0&1\\0&0\end{pmatrix}$, $|\psi\rangle=\scriptsize\begin{pmatrix}0\\1\end{pmatrix}$, $|\phi\rangle=\scriptsize\begin{pmatrix}1\\0\end{pmatrix}$. Then $1=\langle \phi|A|\psi\rangle\neq \overline{\langle\psi|A|\phi\rangle}=0$). But this shows $$ \langle \phi|A|\psi\rangle\neq \overline{\langle\psi|A|\phi\rangle}=I(A|\phi\rangle)|\psi\rangle $$ so $\langle\phi|A\not\equiv I(A|\phi\rangle)$. And indeed, an operator which satisfies $\langle\phi|A= I(A|\phi\rangle)$ for all $\phi\in H$ (and equivalently $\langle \phi|A|\psi\rangle= \overline{\langle\psi|A|\phi\rangle}$ for all $\phi,\psi$) is called hermitian, or self-adjoint in general Hilbert spaces. Hope this helps!