Given a state $|phirangle=frac{1}{sqrt{2}}(|0rangle+e^{itheta}|1rangle)$, how do I know the angle $theta$?

Answer to the first question:

As mentioned in the comments of the question I assume that we can prepare $|phi rangle$ as many as we want. Let's calculate the relative phase for this one qubit pure state:

$$|psi rangle = frac{1}{sqrt{2}} left( |0rangle + e^{itheta}|1rangleright)$$

We are going to execute $2$ different experiments in order to estimate $theta$. In the first experiment we apply this circuit:

circuit_experiment_1.h(q[0])
circuit_experiment_1.measure(q[0], c[0])

The state after Hadamard gate:

$$H frac{1}{sqrt{2}} left( |0rangle + e^{itheta}|1rangleright) = frac{1}{2}left[(1 + e^{itheta})| 0 rangle + (1 - e^{itheta})| 1 rangle right]$$

The probabilities of $|0rangle$ and $|1rangle$ states:

begin{align*} P(0) = frac{1}{4}left| 1 + e^{itheta} right|^2 = frac{1}{2}(1 + cos(theta)) P(1) = frac{1}{4}left| 1 - e^{itheta} right|^2 = frac{1}{2}(1 - cos(theta)) end{align*}

From here we can see that:

$$theta = pm arccosbig(P(0) - P(1)big)$$

because the range of usual principal value arccosine function is equal to $[0, pi]$. So we will need the second experiment in order to estimate the $sign(theta)$. But, before that, how to find $P(0)$ and $P(1)$ with the described experiment? We will need to execute the circuit $N$ times (bigger $N$ gives better precision) and take into accout these relations between measurement outcomes and probabilities:

begin{align*} P(0) = lim_{N rightarrow infty} frac{N_{0}}{N} qquad P(1) = lim_{N rightarrow infty} frac{N_{1}}{N} end{align*}

where $N_{0}$ is the number of $|0rangle$ measurement outcomes and $N_{1}$ is the number of $|1rangle$ measurement outcomes. Also, note that:

$$langle X rangle = langle psi | X | psi rangle = langle psi |H Z H| psi rangle = P(0) - P(1)$$

So, the formula can be written in this way:

$$theta = pm arccos big( langle X rangle big)$$

The sign of the $theta$

Now we should determine the $sign(theta)$ with this circuit:

circuit_experiment_2.sdg(q[0])
circuit_experiment_2.h(q[0])
circuit_experiment_2.measure(q[0], c[0])

The state after applying $S^{dagger}$ and $H$ gates:

$$H S^{dagger} frac{1}{sqrt{2}} left( |0rangle + e^{itheta}|1rangleright) = frac{1}{2}left[(1 - i e^{itheta})| 0 rangle + (1 + i e^{itheta})| 1 rangle right]$$

with the same logic:

begin{align*} P'(0) = frac{1}{4}left| 1 - ie^{itheta} right|^2 = frac{1}{2}(1 + sin(theta)) P'(1) = frac{1}{4}left| 1 + ie^{itheta} right|^2 = frac{1}{2}(1 - sin(theta)) end{align*}

So after determining the $P'(0)$ and $P'(1)$ from the second experiment we will find the sign of the $theta$:

$$sign(theta) = sign(arcsinleft(P'(0) - P'(1)right)) = sign(P'(0) - P'(1))$$

because the range of usual principal value of arcsine function is $[-frac{pi}{2}, frac{pi}{2}]$.

Also, note that for the expectation value of the $Y$ operator (as can be seen from this answer) we have this relation:

$$langle Y rangle = langle psi| Y | psirangle = langle psi| S H Z H S^{dagger} | psirangle = P'(0) - P'(1)$$

By taking this into account and combining two results:

begin{align*} theta = sign big(langle Y rangle big) arccos big(langle X rangle big) end{align*}

An approach for finding the relative phase of an arbitrary pure state is described in this answer.

Answer to the second question:

Here is the circuit for measuring in $M{{({{theta }_{k}})}_{pm }}=left{ 1/sqrt{2}left( |0rangle pm {{e}^{-i{{theta }_{k}}}}|1rangle right) right}$ basis. I assume here that $theta_k$ is given:

circuit.u1(theta_k, q[0])    # q[0] is one of the qubits
circuit.h(q[0])
circuit.measure(q[0], c[0])   #c[0] is a classical bit

If the state was $M(theta _k)_+= 1/sqrt{2}left( |0rangle + e^{-itheta _k}|1rangle right)$, then the outcome of the circuit will be $|0rangle$, and if it was $M(theta _k)_-= 1/sqrt{2}left( |0rangle - e^{-itheta _k}|1rangle right)$, then the outcome of the circuit will be $|1rangle$. So this way we will be able to measure in $M{{({{theta }_{k}})}_{pm }}$ basis.