Yes. If $x_n\in D(L)$ is an ONB of $H$ and $Lx_n=\lambda_n x_n$, then the operator $T$ acting in the obvious way on $D(T)=\{ \sum a_n x_n\in H : \sum \lambda_n^2|a_n|^2<\infty\}$ is self-adjoint. It is also an extension of $L$ because if $x\in D(L)$, then $$ \lambda_n\langle x, x_n\rangle = \langle Lx, x_n \rangle , $$ so $\sum \lambda_n^2|\langle x_n,x\rangle |^2=\|Lx\|^2<\infty$, as required.

(The way I wrote it up this applies to separable Hilbert spaces, but the same argument works in general.)

Yes. If $x_n\in D(L)$ is an ONB of $H$ and $Lx_n=\lambda_n x_n$, then the operator $T$ acting in the obvious way on $D(T)=\{ \sum a_n x_n\in H : \sum \lambda_n^2|a_n|^2<\infty\}$ is self-adjoint. It is also an extension of $L$ because if $x\in D(L)$, then $$ \lambda_n\langle x, x_n\rangle = \langle Lx, x_n \rangle , $$ so $\sum \lambda_n^2|\langle x_n,x\rangle |^2=\|Lx\|^2<\infty$. In the same way, by approximating $(x,Tx)\in G(T)$ by truncated sums $\sum_{n\le N} a_n x_n\in D(L)$, we see that $T=\overline{L}$, so $L$ is essentially self-adjoint, as required.

(The way I wrote it up this applies to separable Hilbert spaces, but the same argument works in general.)

Yes. If $x_n\in D(L)$ is an ONB of $H$ and $Lx_n=\lambda_n x_n$, then the operator $T$ acting in the obvious way on $D(T)=\{ \sum a_n x_n : \sum \lambda_n^2|a_n|^2<\infty\}$ is self-adjoint. It is also an extension of $L$ because if $x\in D(L)$, then $$ \lambda_n\langle x, x_n\rangle = \langle Lx, x_n \rangle , $$ so $\sum \lambda_n^2|\langle x_n,x\rangle |^2=\|Lx\|^2<\infty$, as required.

(The way I wrote it up this applies to separable Hilbert spaces, but the same argument works in general.)

Yes. If $x_n\in D(L)$ is an ONB of $H$ and $Lx_n=\lambda_n x_n$, then the operator $T$ acting in the obvious way on $D(T)=\{ \sum a_n x_n\in H : \sum \lambda_n^2|a_n|^2<\infty\}$ is self-adjoint. It is also an extension of $L$ because if $x\in D(L)$, then $$ \lambda_n\langle x, x_n\rangle = \langle Lx, x_n \rangle , $$ so $\sum \lambda_n^2|\langle x_n,x\rangle |^2=\|Lx\|^2<\infty$, as required.

(The way I wrote it up this applies to separable Hilbert spaces, but the same argument works in general.)