In each of the following cases, we will suppose that one of two single qubit states $\ket{a}$ and $\ket{b}$ is selected randomly with probability $1/2$, but we are not told which. The goal is to guess which state was selected with as high a probability as we can achieve. What will be described is a distinguishing procedure in the form of a unitary operation followed by a measurement in the computational basis, and its success probability.
CASE: (i)
$\ket{a}=\ket{0}$ and $\ket{b}=\frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$
The inner product of the two states $\ket{a}$ and $\ket{b}$ is:
\[\ip{a}{b}=\bra{0}\super{\ket{0}+\ket{1}}=\frac{1}{\sqrt{2}}(\ip{0}{0}+\ip{0}{1})=\frac{1}{\sqrt{2}},\]
since $\ip{i}{j}=\delta_{ij}$ for all $i,j\in\{0,1\}$. Then using the geometrical relation $\ip{a}{b}=\cos{\phi}$, where $\phi\in[0,2\pi)$ denotes the angle between the two states $\ket{a}$ and $\ket{b}$, it is seen that in this case
\[\phi=\arccos{\frac{1}{\sqrt{2}}}=\frac{\pi}{4}.\]
In general, two pure states can only be perfectly distinguished if they are orthogonal. That is, if $\phi=\frac{\pi}{2},\frac{3\pi}{2}$. Therefore, in this case the two states $\ket{a}$ and $\ket{b}$ cannot be perfectly distinguished.
Consider the unitary operation $R$ that rotates states by $\theta=\frac{\pi}{8}$ given by:
\[
R=\begin{pmatrix}
\cos{\frac{\pi}{8}}&-\sin{\frac{\pi}{8}} \\
\sin{\frac{\pi}{8}}&\cos{\frac{\pi}{8}}
\end{pmatrix}
\]
Then applying $R$ to the two computational basis states $\ket{0}$ and $\ket{1}$ gives:
\[\begin{array}{rcl}
R\ket{0}&=&\cos{\frac{\pi}{8}}\ket{0}+\sin{\frac{\pi}{8}}\ket{1}\\
R\ket{1}&=&-\sin{\frac{\pi}{8}}\ket{0}+\cos{\frac{\pi}{8}}\ket{1}
\end{array}\]
Therefore, applying $R$ to the states $\ket{a}$ and $\ket{b}$ yields:
\[\begin{array}{rcl}
R\ket{a}=R\ket{0}&=&\cos{\frac{\pi}{8}}\ket{0}+\sin{\frac{\pi}{8}}\ket{1},\\
R\ket{b}=\frac{1}{\sqrt{2}}R(\ket{0}+\ket{1})
&=&\frac{1}{\sqrt{2}}(\cos{\frac{\pi}{8}}\ket{0}+\sin{\frac{\pi}{8}}\ket{1}-\sin{\frac{\pi}{8}}\ket{0}+\cos{\frac{\pi}{8}}\ket{1})\\
&=&\frac{1}{\sqrt{2}}((\cos{\frac{\pi}{8}}-\sin{\frac{\pi}{8}})\ket{0}+(\cos{\frac{\pi}{8}}+\sin{\frac{\pi}{8}})\ket{1})
\end{array}\]
Since $R$ is unitary the inner product of $R\ket{a}$ and $R\ket{b}$ is preserved, and also the angle $\phi=\frac{\pi}{4}$ between the states is preserved.
When either $R\ket{a}$ or $R\ket{b}$ are measured in the computational basis what will result is one of the basis states $\ket{0}$ or $\ket{1}$, but with certain probability. If the state happened to be $R\ket{a}$, then the probability of observing $\ket{0}$ and $\ket{1}$ is $\cos^2{\frac{\pi}{8}}$ and $\sin^2{\frac{\pi}{8}}$, respectively. If instead the state $R\ket{b}$ was measured, the probability of observing $\ket{0}$ would be $\sin^2{\frac{\pi}{8}}$, and the probability of observing $\ket{1}$ would be $\cos^2{\frac{\pi}{8}}$.
Thus when either $\ket{a}$ or $\ket{b}$ is given uniformly at random, first apply $R$ to the state and then measure. If $\ket{0}$ is observed guess the state $\ket{a}=\ket{0}$, and if the state $\ket{b}$ is observed guess the state $\ket{b}=\frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$.
The probability of guessing the state correctly is given by $\frac{1}{2}(1+\sin{\phi})$, where $\phi=\arccos({\ip{a}{b}})$ is the angle between the states. In this case, with $\phi=\frac{\pi}{4}$, the probability of successfully guessing the correct state is $\frac{1}{2}(1+\sin{\frac{\pi}{4}})\approx .50$.
Case: (ii)
$\ket{a}=\super{\ket{0}+i\ket{1}}$ and $\ket{b}=\super{i\ket{0}+\ket{1}}$
Consider the inner product between $\ket{a}$ and $\ket{b}$:
\[\ip{a}{b}=(\super{\bra{0}-i\bra{1}})(\super{i\ket{0}+\ket{1}})=\frac{1}{2}(i\ip{0}{0}+\ip{0}{1}+\ip{1}{0}-i\ip{1}{1})=\frac{i}{2}-\frac{i}{2}=0\]
Hence, the two states are orthogonal so the angle between them is $\phi=\frac{\pi}{2}$. There must exist a unitary transformation $U$ that maps these two states to the computational basis states $\ket{0}$ and $\ket{1}$. Consider the following unitary map:
\[
U=\frac{1}{\sqrt{2}}\begin{pmatrix}
1&-i \\
i&-1\end{pmatrix},\]
and observe that
\[
U\ket{a}=\frac{1}{\sqrt{2}}\begin{pmatrix}
1&-i \\
i&-1\end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt{2}}\\
\frac{i}{\sqrt{2}}
\end{pmatrix}=
\begin{pmatrix}
1\\
0
\end{pmatrix},\]
\[U\ket{b}=\frac{1}{\sqrt{2}}\begin{pmatrix}
1&-i \\
i&-1\end{pmatrix}
\begin{pmatrix}
\frac{i}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{pmatrix}=
\begin{pmatrix}
0\\
1
\end{pmatrix},
\]
Then by applying the unitary operation $U$ to either of the states $\ket{a}$ or $\ket{b}$ chosen uniformly at random, and then measuring in the computational basis, either the state $\ket{0}$ or $\ket{1}$ will be observed with certainty. If $\ket{0}$ is observed, then the original state was $\ket{a}=\super{\ket{0}+i\ket{1}}$. Otherwise, if the state $\ket{1}$ was observed, then the original state was $\ket{b}=\super{i\ket{0}+\ket{1}}$. Since these states are orthogonal, the probability of success with the procedure is $1$.
Case: (iii)
$\ket{a}=\cos\theta\ket{0}+\sin\theta\ket{1}$ and $\ket{b}=\cos\theta\ket{0}-\sin\theta\ket{1}$, where $\theta\in[0,2\pi)$ is known.
The inner product of the states $\ket{a}$ and $\ket{b}$ is given by:
\[\begin{array}{rcl}\ip{a}{b}&=&(\cos\theta\bra{0}+\sin\theta\bra{1})(\cos\theta\ket{0}-\sin\theta\ket{1})\\
&=&\cos^2{\theta}\ip{0}{0}+\cos{\theta}\sin{\theta}\ip{0}{1}-(\sin{\theta}\cos{\theta}\ip{1}{0}+\sin^2{\theta}\ip{1}{1} \\
&=&\cos^2{\theta}-\sin^2{\theta} \\
&=&\cos{2\theta}
\end{array}\]
Therefore the angle between the two states is $\phi=2\theta$. If $\theta=0$, then $\phi=2\theta=0$ and the states are actually the same. In this case, there is no improved way of distinguishing between the two states besides merely guessing with a $50/50$ chance. Otherwise, consider the unitary operator $R$ that performs a rotation by $\pi/4$:
\[R=\begin{pmatrix}\cos{\frac{\pi}{4}}&-\sin{\frac{\pi}{4}} \\
\sin{\frac{\pi}{4}}&\cos{\frac{\pi}{4}}
\end{pmatrix}\]
Geometrical considerations show that applying $R$ to the states $\ket{a}$ and $\ket{b}$ yields:
\[\begin{array}{rcl}R\ket{a}&=&\cos{(\theta-\pi/4)}\ket{0}+\sin{(\theta-\pi/4)}\ket{1},\\
R\ket{b}&=&\cos{(\theta+\pi/4)}\ket{0}+\sin{(\theta+\pi/4)}\ket{1}.
\end{array}\]
Then if $R\ket{a}$ is measured in the computation basis,$\ket{0} $will be observed with probability $\cos^2{(\theta-\pi/4)}$ and $\ket{1}$ will be observed with probability $\sin^2{(\theta-\pi/4)}$. On the other hand, if the state $R\ket{b}$ is measured, the states $\ket{0}$ and $\ket{1}$ will be observed with probability $\cos^2{(\theta+\pi/4)}$ and $\sin^2{(\theta+\pi/4)}$, respectively.
Therefore, for $0\leq\theta\leq\pi/2$ or $\pi\leq\theta\leq3\pi/2$, if the state $\ket{0}$ is observed, guess that the original state was $\ket{a}$. For $\pi/2\leq\theta\leq\pi/2$ or $3\pi/2\leq\theta\leq2\pi$, if $\ket{1}$ is observed, then guess that the original state was $\ket{b}$. This method has a probability of success given by $P=\frac{1}{2}(1+\sin{(2\theta)})$.