Matrix rows or columns are traditionally listed under $(x,y,z)$ order.
Cyclically change the pairs under consideration i.e $(x,y)\to(y,z)\to(z,x)$. The pairs $(x,y)$ and $(y,z)$ show up in the same order in the matrix but the $(z,x)$ shows up in reverse in the matrix. That is the cause of apparent discrepancy but really there is no discrepancy.
For example write
$x'=x\cos \alpha - y \sin \alpha$
$y'=x\sin \alpha + y \cos \alpha$
now change $(x,y)\to(y,z)\to(z,x)$ and $\alpha\to \beta \to \gamma$ and write the three matrices to see how $(z,x)$ part gets flipped.
Edit:
If you want them to look alike then give up the matrix notation and instead write
$y'=y\cos \beta - z \sin \beta$
$z'=y\sin \beta + z \cos \beta$
And
$z'=z\cos \gamma - x \sin \gamma$
$x'=z\sin \gamma + x \cos \gamma$
In each instance if you try to write $\left[ \matrix{ x' \cr y' \cr z'}\right]$ in terms of $\left[ \matrix{ x \cr y \cr z}\right]$ you will see that the mystery goes away.
$$
\begin{bmatrix}
\cos t&-\sin t&|&1&0\\ \sin t&\cos t&|&0&1
\end{bmatrix}
\xrightarrow{\frac1{\cos t}R1}
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ \sin t&\cos t&|&0&1
\end{bmatrix}
\xrightarrow{R2-\sin t\,R1}
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&\cos t+\frac{\sin^2t}{\cos t}&|&-\frac{\sin t}{\cos t}&1
\end{bmatrix}=
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&\frac1{\cos t}&|&-\frac{\sin t}{\cos t}&1
\end{bmatrix}
\xrightarrow{\cos t\,R2}
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&1&|&-\sin t&\cos t
\end{bmatrix}
\xrightarrow{R1+\frac{\sin t}{\cos t}R2}
\begin{bmatrix}
1 &0&|&\frac1{\cos t}-\frac{\sin^2t}{\cos t}&\sin t\\ 0&1&|&-\sin t&\cos t
\end{bmatrix}
=\begin{bmatrix}
1 &0&|&\cos t&\sin t\\ 0&1&|&-\sin t&\cos t
\end{bmatrix}
$$
This is a terrible method to calculate the inverse of any $2\times 2$ matrix.
Edit: of course this does not work when $\cos t=0$; but this is a much easier case: you simply divide by $\sin t$ and permute the rows.
Best Answer
Hint:
(a) $\quad$ Calculate $R(-\theta)R(\theta)$
(c) $\quad$ By definition: symmetric when $R(\theta)^T=R(\theta)$
$\quad \quad \,$ By $(b)$ then $R(\theta)=R(-\theta) \implies \sin \theta=-\sin \theta$