= Solution
The <spectrum of a bounded operator> is $\sigma(T)=\{\lambda\in\mathbb C:T-\lambda I\text{ has no bounded everywhere-defined inverse}\}$. Initially define $T$ on finite linear combinations of the <orthonormal basis>. The <bilateral shift operator> $Se_n=e_{n+1}$ preserves the <Hilbert space> norm and extends to a unitary operator on $H$; its inverse is $S^*e_n=e_{n-1}$. Thus $T=S+S^*$ extends uniquely by density, is bounded with norm at most two, and is <self-adjoint>.
For $v_N=(2N+1)^{-1/2}\sum_{n=-N}^Ne_n$, $\|v_N\|=1$ and the only nonzero coefficients of $(T-2I)v_N$ occur at $-N-1,-N,N,N+1$. Hence
$$
\|(T-2I)v_N\|^2=\frac4{2N+1}\longrightarrow0.
$$
It follows that $\|Tv_N\|\to2$, proving $\boxed{\|T\|=2}$. Also $\boxed{2\in\sigma(T)}$: a bounded inverse of $T-2I$ would give $1\leq\|(T-2I)^{-1}\|\,\|(T-2I)v_N\|\to0$. Thus two belongs to the <approximate point spectrum>.
An <eigenvector> $x=\sum x_ne_n$ would satisfy $\lambda x_n=x_{n-1}+x_{n+1}$. Self-adjointness and the norm bound force $\lambda\in[-2,2]$. If $|\lambda|<2$, write $\lambda=2\cos\theta$ with $0<\theta<\pi$; the general complex solution of the <linear recurrence relation> is
$$
x_n=Ae^{in\theta}+Be^{-in\theta}.
$$
The Cesàro mean of $|x_n|^2$ over $n=-N,\ldots,N$ tends to $|A|^2+|B|^2$, since the mixed geometric sum divided by $2N+1$ tends to zero. For an $\ell^2$ sequence the same mean tends to zero. Therefore $A=B=0$. At $\lambda=\pm2$, the general solution is $(A+Bn)(\pm1)^n$, again square summable only when zero. This also covers $\lambda=0$ through $\theta=\pi/2$. Thus $\boxed{T\text{ has no eigenvectors}}$.
Finally the bounded sequence $e_{3j}$ has images $Te_{3j}=e_{3j-1}+e_{3j+1}$ with pairwise disjoint supports. Distinct images have distance two, so no subsequence converges. Hence $\boxed{T\text{ is not compact}}$.
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