$\newcommand{\la}{\lambda}\newcommand{\R}{\mathbb R}$For any $x=(x_1,\dots,x_n)\in\R^n$ we have $Tx=(x_2,\dots,x_n,x_1)$ and $Ax=(x_3+x_4,0\dots,0)$, so that for $U:=T+A$ we have $Ux=(x_2+x_3+x_4,x_3,\dots,x_n,x_1)$.

Let now $(x_1,\dots,x_n)\in\R^n$ be an eigenvector of $U$ belonging to an eigenvalue $\la$ of $U$. Then \begin{equation} \left\{ \begin{aligned} &x_k=\la x_{k-1}\text{ for }k=3,\dots,n, \\ &x_2+x_3+x_4=\la x_1, \\ &x_1=\la x_n. \end{aligned} \right. \end{equation} So, $x_k=\la^{k-2} x_2$ for $k=3,\dots,n$, $x_1=\la x_n=\la^{n-1} x_2$, $(1+\la+\la^2)x_2=\la x_1=\la^n x_2$. So, $x_2\ne0$ and without loss of generality (wlog) $x_2=1$. So, $\la$ is one of the $n$ roots $\la_1,\dots,\la_n$ of the equation $1+\la+\la^2=\la^n$, and the eigenvector $(x_{j,1},\dots,x_{j,n})$ belonging to the eigenvalue $\la_j$ is given by th formulas \begin{equation} x_{j,2}=1,\quad x_{j,1}=\la_j^{n-1},\quad x_{j,k}=\la_j^{k-2}\text{ for }k=3,\dots,n. \end{equation}

$\newcommand{\la}{\lambda}\newcommand{\R}{\mathbb R}$For any $x=(x_1,\dots,x_n)\in\R^n$ we have $Tx=(x_2,\dots,x_n,x_1)$ and $Ax=(x_3+x_4,0\dots,0)$, so that for $U:=T+A$ we have $Ux=(x_2+x_3+x_4,x_3,\dots,x_n,x_1)$.

Let now $(x_1,\dots,x_n)\in\R^n$ be an eigenvector of $U$ belonging to an eigenvalue $\la$ of $U$. Then \begin{equation} \left\{ \begin{aligned} &x_k=\la x_{k-1}\text{ for }k=3,\dots,n, \\ &x_2+x_3+x_4=\la x_1, \\ &x_1=\la x_n. \end{aligned} \right. \end{equation} So, $x_k=\la^{k-2} x_2$ for $k=3,\dots,n$, $x_1=\la x_n=\la^{n-1} x_2$, $(1+\la+\la^2)x_2=\la x_1=\la^n x_2$. So, $x_2\ne0$ and without loss of generality (wlog) $x_2=1$. So, $\la$ is one of the $n$ roots $\la_1,\dots,\la_n$ of the equation $1+\la+\la^2=\la^n$, and the eigenvector $(x_{j,1},\dots,x_{j,n})$ belonging to the eigenvalue $\la_j$ is given by th formulas \begin{equation} x_{j,2}=1,\quad x_{j,1}=\la_j^{n-1},\quad x_{j,k}=\la_j^{k-2}\text{ for }k=3,\dots,n. \end{equation}

It is now shown (see this question and the answers to it) that the eigenvalues $\la_1,\dots,\la_n$ are pairwise distinct and hence the corresponding eigenvectors form a basis of $\R^n$.

$\newcommand{\la}{\lambda}\newcommand{\R}{\mathbb R}$Let $(x_1,\dots,x_n)\in\R^n$ be an eigenvector of $U:=T+A$ belonging to an eigenvalue $\la$. Then \begin{equation} \left\{ \begin{aligned} &x_k=\la x_{k-1}\text{ for }k=3,\dots,n, \\ &x_2+x_3+x_4=\la x_1, \\ &x_1=\la x_n. \end{aligned} \right. \end{equation} So, $x_k=\la^{k-2} x_2$ for $k=3,\dots,n$, $x_1=\la x_n=\la^{n-1} x_2$, $(1+\la+\la^2)x_2=\la x_1=\la^n x_2$. So, $x_2\ne0$ and without loss of generality (wlog) $x_2=1$. So, $\la$ is one of the $n$ roots $\la_1,\dots,\la_n$ of the equation $1+\la+\la^2=\la^n$, and the eigenvector $(x_{j,1},\dots,x_{j,n})$ belonging to the eigenvalue $\la_j$ is given by th formulas \begin{equation} x_{j,2}=1,\quad x_{j,1}=\la_j^{n-1},\quad x_{j,k}=\la_j^{k-2}\text{ for }k=3,\dots,n. \end{equation}

$\newcommand{\la}{\lambda}\newcommand{\R}{\mathbb R}$For any $x=(x_1,\dots,x_n)\in\R^n$ we have $Tx=(x_2,\dots,x_n,x_1)$ and $Ax=(x_3+x_4,0\dots,0)$, so that for $U:=T+A$ we have $Ux=(x_2+x_3+x_4,x_3,\dots,x_n,x_1)$.

Let now $(x_1,\dots,x_n)\in\R^n$ be an eigenvector of $U$ belonging to an eigenvalue $\la$ of $U$. Then \begin{equation} \left\{ \begin{aligned} &x_k=\la x_{k-1}\text{ for }k=3,\dots,n, \\ &x_2+x_3+x_4=\la x_1, \\ &x_1=\la x_n. \end{aligned} \right. \end{equation} So, $x_k=\la^{k-2} x_2$ for $k=3,\dots,n$, $x_1=\la x_n=\la^{n-1} x_2$, $(1+\la+\la^2)x_2=\la x_1=\la^n x_2$. So, $x_2\ne0$ and without loss of generality (wlog) $x_2=1$. So, $\la$ is one of the $n$ roots $\la_1,\dots,\la_n$ of the equation $1+\la+\la^2=\la^n$, and the eigenvector $(x_{j,1},\dots,x_{j,n})$ belonging to the eigenvalue $\la_j$ is given by th formulas \begin{equation} x_{j,2}=1,\quad x_{j,1}=\la_j^{n-1},\quad x_{j,k}=\la_j^{k-2}\text{ for }k=3,\dots,n. \end{equation}