They both represent the same state since they differ only in a global phase of -1.

Same as (a), the global phase is $\i$.

Same as (a) and (b) Global phase is -1

Given two orthogonal basis $\ket k$ and $\ket{k^\perp}$. The qubit $\ket \psi$ can be seen as $\ket \psi = \alpha\ket k + \beta\ket{k^\perp}$

Where $\alpha$ and $\beta$ are the projection of $\ket \psi$ in $\ket k$ and $\ket{k^\perp})$ respectively.

The probability to measure $\ket k$ is ${\mid \alpha\mid}^2$ and to measure $\ket {k^\perp}$ is ${\mid \beta\mid}^2$ and ${\mid \alpha\mid}^2 + {\mid \beta\mid}^2 = 1$.

Given another basis state $\ket v$ . The angle $\theta$ between state $\ket v$ and $\ket \psi$ is given by:

(1)The projection of $\ket \psi$ in $\ket v$ is given by $cos(\theta).{\mid v\mid . \mid \psi\mid}$, therefore the probability $P_\psi$ to measure $\ket v$ is equal to

(2)Since:

(3)where $a_j^*$ is the complement of the complex number $a_j$

Then

(6)$\ket \psi =(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ and $\ket v = (\frac{1}{\sqrt{2}},\frac{i}{\sqrt{2}})$ then $\bra v = (\frac{1}{\sqrt{2}},\frac{-i}{\sqrt{2}})$

(1)$\ket \psi =(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ and $\ket v = (\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}})$ then $\bra v = (\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}})$

(1)both states are orthogonal

$\ket \psi =(\frac{1}{\sqrt{2}},\frac{i}{\sqrt{2}})$ and $\ket v = (\frac{-1}{\sqrt{2}},\frac{i}{\sqrt{2}})$ then $\bra v = (\frac{-1}{\sqrt{2}},\frac{-i}{\sqrt{2}})$

(1)$\frac{1}{\sqrt{2}}(\ket + + \ket -) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) + \frac{1}{\sqrt{2}}(\ket 0 - \ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + \ket 1) + \frac{1}{2}(\ket 0 - \ket 1)$

$\frac{1}{\sqrt{2}}(\ket + + \ket -) = \ket 0$

Both are the same state with global phase 1

$\frac{1}{\sqrt{2}}(\ket i - \ket{-i}) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + i\ket 1) - \frac{1}{\sqrt{2}}(\ket 0 - i\ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + i\ket 1) - \frac{1}{2}(\ket 0 - i\ket 1)$

$\frac{1}{\sqrt{2}}(\ket i - \ket{-i}) = i\ket 1$

Both are the same state with global phase $-i$

$\frac{1}{\sqrt{2}}(\ket i + \ket{-i}) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + i\ket 1) + \frac{1}{\sqrt{2}}(\ket 0 - i\ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + i\ket 1) + \frac{1}{2}(\ket 0 - i\ket 1)$

$\frac{1}{\sqrt{2}}(\ket i + \ket{-i}) = \ket 0$

por 2.2 g) $\Rightarrow \frac{1}{\sqrt{2}}(\ket + + \ket -) = \ket 0$

Both are the same state with global phase 1