I need to find the eigenvalues and vector for the operator in hilbert space $H$:
$$Tx=\langle x,v\rangle w+\langle x,w\rangle v,$$
where $\langle v,w \rangle =0$, but not necessarily with norm=1.
I already checked that it is compact and self adjoint so there must be eigenvalues
I tried the form of $\alpha w+\beta v$ which works only if $\alpha =\beta$ and if both $w$ and $v$ are normalize, and they are not
I will assume $v, w \neq 0$. The $v = 0$ and/or $w = 0$ cases are easier, and can be decided separately. Note then that $\langle v, w \rangle = 0$ implies $v$ and $w$ are linearly independent.
Using my hint in the comments, I do believe we can indeed find the other eigenvectors (other than the ones in the kernel, i.e. the vectors perpendicular to both $v$ and $w$). Let $u = \alpha v + \beta w$, and assume it is an eigenvector with respect to $\lambda$. As $v, w$ are perpendicular, we have $$\langle u, v\rangle = \langle \alpha v + \beta w, v\rangle = \alpha \|v\|^2,$$ and similarly $\langle u, w\rangle = \beta\|w\|^2$. We then have, by the linear independence of $v$ and $w$, \begin{align*} T(u) - \lambda u = 0 &\iff \langle u, v\rangle w + \langle u, w \rangle v - \lambda(\alpha v + \beta w) = 0 \\ &\iff (\alpha\|v\|^2 - \lambda \beta)w + (\beta\|w\|^2 - \lambda\alpha)v = 0 \\ &\iff \begin{cases} \alpha\|v\|^2 - \lambda \beta = 0 \\\beta\|w\|^2 - \lambda \alpha = 0 \end{cases} \\ &\iff \begin{cases} \alpha\|v\|^2 \|w\|^2 - \lambda \beta \|w\|^2 = 0 \\\beta\|w\|^2 = \lambda \alpha \end{cases} \\ &\iff \begin{cases} \alpha\|v\|^2 \|w\|^2 - \lambda^2 \alpha = 0 \\\beta\|w\|^2 = \lambda \alpha \end{cases}. \end{align*} From here, we can deduce that $\alpha = 0$ or $\lambda = \pm \|v\|\|w\|$. If $\alpha = 0$ were true, then $\beta\|w\|^2 - \lambda \alpha$ would imply $\beta = 0$ too, in which case we don't have an eigenvector. Thus, the only possible eigenvalues are $\lambda = \pm \|v\|\|w\|$.
We now need to confirm that these are not just possible eigenvalues, but actual eigenvalues, by finding actual eigenvectors. If $\lambda = \|v\|\|w\|$, then $\beta\|w\|^2 - \lambda \alpha = 0$ implies $$\beta\|w\|^2 = \|v\|\|w\|\alpha \implies \alpha = \beta\frac{\|w\|}{\|v\|}.$$ We are not going to get better than this, as $\alpha$ and $\beta$ can be scaled by any non-zero scalar, and we will still get an eigenvector. So, I will choose $\alpha = \|w\|$ and $\beta = \|v\|$. This gives us the eigenvector $u = \|w\|v + \|v\|w$. We can check this: \begin{align*} T(\|w\|v + \|v\|w) &= \langle \|w\|v + \|v\|w, v\rangle w + \langle \|w\|v + \|v\|w, w \rangle v \\ &= \|w\|\langle v, v\rangle w + \|v\|\langle w, w\rangle v \\ &= \|v\|\|w\|(\|v\|w + \|w\|v), \end{align*} as required. Thus, we have an eigenvector, and so $\|v\|\|w\|$ is definitely an eigenvalue of $T$.
I'll leave the $-\|v\|\|w\|$ case to you.