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2

The reason is that state $H^{\otimes n} |0\rangle^ {\otimes n} = \frac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1}|i\rangle$, where $|i\rangle$ being a binary representation of decimal number $i$, is a ground state of Hamiltonian $$\mathcal{H}_0 =\sum_{i=1}^{2^n} \sigma_i ^x,$$ where $\sigma_i ^x$ is $X$ gate applied on $i$th qubit whereas identity gate is applied on ...

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A maximally entangled state is a state that has maximum mutual information of the random variables. I think you should first clear your concept about entanglement. http://www.cmi.ac.in/~neelraha/Resources/Internships/MayJuly2016/Maximally_Entangled_States.pdf

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I submitted an issue on GitHub here.

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"As far as I understand there aren't many rigorous results on performance of these algorithms, similar to many classic machine learning approaches." You are correct in that, unlike Grover's algorithm where we can prove that a search that would cost $\mathcal{O}(N)$ on a classical computer can be done with only $\mathcal{O}(\sqrt{N})$ on a quantum ...

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You can have a look at how the QAOAAnsatz class is used in the tests here. Note that the link given @KAJ226 in the question comments does not use QAOAAnsatz directly but use the QAOA class that forward the given operator to the QAOAAnsatz class. So when the line result = qaoa.compute_minimum_eigenvalue(qubit_op) in cell 6 of KAJ226 link is executed, it will ...

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The QAOAAnsatz was built for the QAOA algorithm, which if you look at the qaoa code in Qiskit you will see it builds a QAOAAnsatz instance internally. Hopefully looking at that helps you use it https://github.com/Qiskit/qiskit-terra/blob/5ca967557b21828c0760763b7f0c5870e5f032d9/qiskit/algorithms/minimum_eigen_solvers/qaoa.py#L131-L132

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