Solution of Part 1
From class we know that, an arbitrary single qubit unitary operator can be written as
(1)\[
U = \exp\left[-\frac{it}{\hbar}(aI+bX+cY+dZ)\right], a,b,c,d\in \mathbb{R}
\]
Also, for operator \(A\) and \(B\), if we have \([A, B] = 0\), then we have
(2)\[
\exp[A+B] = \exp(A)\exp(B)
\]
Since the identity matrix commutes with any matrices, we could re-write eq. (1) as
(3)\[\begin{split}
\begin{align}
U &= \exp\left[-\frac{it}{\hbar}(aI+bX+cY+dZ)\right] \\
&= \exp\left[-\frac{iat}{\hbar}I\right]\exp\left[-\frac{it}{\hbar}(bX+cY+dZ)\right]
\end{align}
\end{split}\]
Let \(l = \sqrt{b^2+c^2+d^2}\), we could re-write eq. (3) as
(4)\[\begin{split}
\begin{align}
U &= \exp\left[-\frac{iat}{\hbar}I\right]\exp\left[-\frac{it}{\hbar}(bX+cY+dZ)\right] \\
&= \exp\left[-\frac{iat}{\hbar}\right]\exp\left[-\frac{ilt}{\hbar}\left(\frac{b}{l}X+\frac{c}{l}Y+\frac{d}{l}Z\right)\right] \\
&= \exp\left[-\frac{iat}{\hbar}\right]\exp\left[-\frac{ilt}{\hbar}\left(n_xX+n_yY+n_zZ\right)\right] \\
\end{align}
\end{split}\]
where we have \(\exp(-{iat}I/{\hbar}) = \exp(-{iat}/{\hbar})\), and write \(n_x = b/l,n_y = c/l,n_z = d/l\) and \((n_x, n_y, n_z)\) is naturally a unit vector. Compare with the definition of \(R_{\hat{n}}(\theta) = \exp(-i\theta\hat{n}\cdot\vec{\sigma}/2)\), we could re-write eq. (4) as
(5)\[\begin{split}
\begin{align}
U &= \exp\left[-\frac{iat}{\hbar}\right]\exp\left[-\frac{ilt}{\hbar}\left(n_xX+n_yY+n_zZ\right)\right] \\
&= \exp\left[-i\cdot\frac{at}{\hbar}\right]\exp\left[-i\frac{2lt}{\hbar}\cdot\frac{\hat{n}\cdot\vec{\sigma}}{2}\right]\\
&= \exp(i\alpha)R_{\hat{n}}(\theta)
\end{align}
\end{split}\]
where \(\alpha = -at/\hbar\) and \(\theta = 2lt/\hbar\).
Solution of Part 2
The Hadamard gate can be written in the form \(H = {X+Z}/{\sqrt{2}}\), then we could start from the assumption that \(n_x = n_z = 1/\sqrt{2}\). In this case, we have
(6)\[
H = \exp(i\alpha)R_{\hat{n}}(\theta) = \exp(i\alpha)\left[\cos \frac{\theta}{2}I - i\sin\frac{\theta}{2}(n_xX + n_yY + n_zZ)\right]
\]
If \(\theta = \pi\), we could write eq. (6) as
(7)\[
H = \exp(i\alpha)R_{\hat{n}}(\theta) = \exp(i\alpha)\left[- i(n_xX + n_yY + n_zZ)\right]
\]
Then we could find \(\exp(i\alpha) = i\) and \(\alpha = \pi\).
Solution of Part 3
For \(\hat{n} = \hat{z} = (0,0,1)\), we could write the phase gate as
(8)\[\begin{split}
\begin{align}
S = \begin{pmatrix}
1 & 0 \\ 0 & i
\end{pmatrix} &= \exp(i\alpha)R_{\hat{z}}(\theta)= \exp(i\alpha)\begin{pmatrix}
e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2}
\end{pmatrix} = \begin{pmatrix}
e^{i(\alpha - \theta/2)} & 0 \\ 0 & e^{i(\alpha + \theta/2)}
\end{pmatrix}
\end{align}
\end{split}\]
In order to find \((\alpha, \theta)\), we need to find the solution of below equations,
(9)\[\begin{split}
\begin{cases}
e^{i(\alpha - \theta/2)} = 1\\
e^{i(\alpha + \theta/2)} = i
\end{cases} \iff \begin{cases}
\alpha - \theta/2 = 0\\
\alpha + \theta/2 = \pi/2
\end{cases}
\end{split}\]
Solve the equation, we could find \(\alpha = \pi/4\) and \(\theta = \pi/2\). Then
(10)\[\begin{split}
S =\exp(i\pi/4)R_{\hat{z}}(\pi/2) =\begin{pmatrix}
1 & 0 \\ 0 & i
\end{pmatrix}
\end{split}\]