Since $A$ is orthogonal, we know that $AA^T=I$, that is $A^T=A^{-1}$. This implies that $\det(A)=\pm1$. We are given that $\det(A)=1$.
An orthogonal matrix represents an isometry; that is, $ |Ax|^2=x^TA^TAx=x^Tx=|x|^2 $ There are two kinds of isometries in $\mathbb{R}^3$: rotations and reflections. rotations have determinant $1$ (they preserve orientation) and reflections have determinant $-1$ (they reverse orientation). Thus, $A$ is a rotation.
Rotations by an angle other than $0$ or $\pi$, have one real eigenvector, the axis of rotation, and it has eigenvalue $1$. Let $v$ be the axis of rotation of $A$. That is, $Av=v$. Multiplying by $A^{-1}$ yields $v=A^{-1}v=A^Tv$. Therefore, $v$ is also an eigenvector of $A^T$ with eigenvalue $1$. Thus, $ Mv=(A-A^T)v=v-v=0 $ Suppose that $Mu=0$, then $Au=A^Tu$. Multiplying by $A$ yields $A^2u=u$. Thus, $ Mu=0\implies(A+I)(A-I)u=0 $ Since the only real eigenvalue of $A$ is $1$, $A+I$ is invertible, which implies that $(A-I)u=0$. That is, $u$ is parallel to the axis of rotation. Thus, the nullity of $M$ is $1$ and its rank is $3-1=2$.
To answer the second part of the question, we first notice that we are looking for an eigenvector with eigenvalue $1$. That is a $v$ so that $ \begin{bmatrix} a_{11}-1&a_{12}&a_{13}\\ a_{21}&a_{22}-1&a_{23}\\ a_{31}&a_{32}&a_{33}-1\\ \end{bmatrix} v=0 $ To get a vector perpendicular to the first two rows of $A$, we can take their cross product: $ v=\begin{bmatrix} a_{12}a_{23}-(a_{22}-1)a_{13}\\ a_{13}a_{21}-(a_{11}-1)a_{23}\\ (a_{11}-1)(a_{22}-1)-a_{21}a_{12} \end{bmatrix} $ Since the rows of $A-I$ are linearly dependent, this $v$ is perpendicular to the third row as well. If the previous vector is $0$ (e.g. if the first two rows of $A-I$ are parallel), then try the cross product of the second and third rows of $A-I$: $ v=\begin{bmatrix} (a_{22}-1)(a_{33}-1)-a_{32}a_{23}\\ a_{23}a_{31}-a_{21}(a_{33}-1)\\ a_{21}a_{32}-a_{31}(a_{22}-1) \end{bmatrix} $ or the cross product of the third and first rows of $A-I$: $ v=\begin{bmatrix} a_{32}a_{13}-a_{12}(a_{33}-1)\\ (a_{33}-1)(a_{11}-1)-a_{31}a_{13}\\ a_{31}a_{12}-(a_{11}-1)a_{32} \end{bmatrix} $