Yes. The eigenstates of the (non-hermitian) operator
$$
\hat O=\hat \alpha+i\lambda\hat \beta
$$
are the so-called intelligent states. They were introduced in
Aragone, C., Guerri, G., Salamo, S. and Tani, J.L., 1974. Intelligent spin states. Journal of Physics A: Mathematical, Nuclear and General, 7(15), p.L149.
for angular momentum as states that saturate the uncertainly relation, i.e. for those eigenstate $\vert\lambda\rangle$ one has
$$
\Delta \alpha^2\Delta \beta^2=\frac{1}{4}\vert\langle \lambda\vert [\hat \alpha,\hat \beta]\vert \lambda\rangle\vert^2\, ,
$$ where the $=$ sign replaces the $\ge$ sign.
There are several ways of constructing intelligent states for various types of systems. A typical example of paper on the topic is
El Kinani, A.H. and Daoud, M., 2002. Generalized coherent and intelligent states for exact solvable quantum systems. Journal of Mathematical Physics, 43(2), pp.714-733.
but searching for "intelligent states" in GoogleScholar will return a good number of hits.
The coherent states are examples of intelligent states for $\hat x$ and $\hat p$, and indeed the squeezed states for $\hat x$ and $\hat p$ are also "intelligent". In general, since the operator $\hat O$ is not hermitian, intelligent states are not orthogonal, and again the coherent states or squeezed states are examples of non-orthogonal states. Since the operator $\hat O$ is not hermitian, its eigenvalues can be complex.
Intelligent spin states have applications in quantum optics, as discussed for instance in
Wodkiewicz, K. and Eberly, J.H., 1985. Coherent states, squeezed fluctuations, and the SU (2) am SU (1, 1) groups in quantum-optics applications. JOSA B, 2(3), pp.458-466.
Now, using this bra-ket notation we can compute the inner product of some operator, say $\hat{H}$, so $\langle\psi|\hat{H}|\psi\rangle$ defines the eigenvalue of some hermitian operator $\hat{H}$.
The inner product is a thing between two vectors - "the inner product of some operator" is not a meaningful phrase. If $|\psi\rangle$ is a normalized eigenvector of $\hat H$ with eigenvalue $\lambda$, then it's true that $\langle \psi|\hat H|\psi\rangle = \lambda$, but the definition of an eigenvector/eigenvalue pair is that $\hat H|\psi\rangle = \lambda|\psi\rangle$.
This is also called the expectation value of $\hat{H}$ and describes the probability of measuring this operator given the state $\psi$.
$\langle \psi|\hat H|\psi\rangle$ is referred to as the expectation value (or expected value) of $\hat H$ (corresponding to the normalized state vector $|\psi\rangle$). The interpretation of this number is that if you take a large number of identical systems all prepared in the state $|\psi\rangle$ and measured $\hat H$ in each of them, you would expect the mean value of all of those results to be $\langle \psi|\hat H|\psi\rangle$.
We can also derive the inner product $\langle\phi|\psi\rangle$. I must admit that I am a little confused about this representation although it makes sence mathematically. Does this mean the probability of being in the state $\phi$ given the state $\psi$?
There is no immediate physical interpretation of the inner product between two vectors - it is a quantity which shows up in all kinds of different contexts, and essentially measures the "overlap" between $\psi$ and $\phi$. It is analogous to the ordinary dot product between vectors in $\mathbb R^3$.
If $\psi$ is a normalized state vector representing the state of the system and $\phi$ is a normalized eigenvector of some observable $\hat A$ with (non-degenerate) eigenvalue $\lambda$, then $|\langle \phi|\psi\rangle|^2$ is the probability of measuring $\hat A$ to take the value $\lambda$. So that is one context in which the expression could arise. But trying to assign a single physical meaning to the inner product is like trying to assign a single physical meaning to the dot product between vectors in $\mathbb R^3$.
Best Answer
This is a common misconception. Performing a measurement on a system is not mathematically represented by applying the corresponding Hermitian operator to that state. The operator just tells us the possible measurement outcomes through its eigenvalues, and it allows us to determine the probability of measuring each of those values by expressing our state in the eigenbasis of the operator. $\hat A|\psi\rangle$ will not give you the measurement outcome.
Uncertainty and error are two different things in the context of Quantum Mechanics. For a given state we can compute the uncertainty of a measurement as you have described. What this means is that if we were to prepare a bunch of similar states and measure the observable in question, we would find the standard deviation of those measurements to (ideally) have a value of $\Delta A_\psi$. Error would come into play in the method of measurement, where the returned values have some error associated with them. Uncertainty is "baked into" Quantum Mechanics and is independent of the method of measurement, whereas error depends on how the measurement was done. A system can have no uncertainty with respect to an observable yet the measurements will still have non-zero error.
To address another confusion, uncertainty is not the same thing as the uncertainty principle, which relates the uncertainties of two observables. For a given state, we can determine the uncertainty of a single observable. Or, using the language from above, we can determine the standard deviation of measurements of an observable from similarly prepared states.