How to calculate the probability of a particular value of an observable being measured
$$
\newcommand{\ket}[1]{\,\lvert{#1}\rangle}
\newcommand{\bra}[1]{\langle{#1}\rvert\,}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
\newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
\newcommand{\op}[1]{\hat{#1}}
$$

Solution should be a pretty straightforward application of the Born rule.



In this particular case, we have a self-adjoint operator $\op{L_z}$ with eigenfunctions and eigenvalues satisfying 
$$
  \op{L_z} \psi_l(\theta) = l \hbar \psi_l(\theta) \, ,
  \quad \text{where} \quad 
  \psi_l(\theta) = \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{\mathrm{i} l \theta}
  \quad \text{and} \quad 
  l = 0, \pm 1, \pm 2, \dotsc
$$
such that any wave function $\psi$ can be expanded over orthonormal basis $\{ \psi_l(\theta) \}_{l=-\infty}^{\infty}$ of the corresponding eigenfunctions of $\op{L_z}$ as follows,
$$
  \psi(\theta)
  =
  \sum\limits_{l=-\infty}^{\infty} c_l \psi_l(\theta) \, ,
  \quad \text{where} \quad
  c_l 
  = 
  \braket{\psi_l}{\psi}
  =
  \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta \, .
$$
And a measurement on a system in arbitrary state $\psi$ can yield any of eigenvalues $l \hbar$ with the probability given as follows
$$
  \Pr(l \hbar) = |c_l|^{2} = c_l^* c_l \, .
$$



Now, since $\psi(\theta)$ is defined to be
$$
  \psi(\theta)
  =
  \begin{cases} 
  \frac{1}{\sqrt{\pi}} &amp; \text{if} \quad \theta \in [0, \pi] \\
  0            &amp; \text{if} \quad \theta \in [\pi, 2 \pi]
  \end{cases} \, ,
$$
the coefficient $c_l$ is equal to
$$
  c_l 
  =
  \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta
  =
  \int\limits_{0}^{\pi} \psi_l^*(\theta) 1/\sqrt{\pi} \mathrm{d} \theta
  +
  \int\limits_{\pi}^{2 \pi} \psi_l^*(\theta) 0 \mathrm{d} \theta \, ,
$$
where the second term trivially vanishes leading to
$$
  c_l 
  =
  \int\limits_{0}^{\pi}
  \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{-\mathrm{i} l \theta} \frac{1}{\sqrt{\pi}}
  \mathrm{d} \theta
  =
  \int\limits_{0}^{\pi}
  \frac{1}{\sqrt{2} \pi} \mathrm{e}^{-\mathrm{i} l \theta}
  \mathrm{d} \theta \, .
$$
And then some simple algebra leads to the final answer, which I leave to the OP, since it is far into the night for me and I'm afraid I will make a lot of stupid mistakes.