Are the following linear operators well defined bounded operators $\ell^2 \to \ell^1$: $$\begin{align*} T: (x_k)_{k=1}^\infty &\mapsto (k^{-1}x_{k+2})_{k=1}^\infty \\\\ L:(x_k)_{k=1}^\infty & \mapsto (x_8+k^{-4}x_k)_{k=1}^\infty\end{align*}$$
What does it mean to be well defined in this context? I think that the first one is not neccessarily bounded as $$\sum_{k=1}^\infty |\frac{1}{k}x_{k+2}| \le \left( \sum_{k=1}^\infty |\frac{1}{k^2}| \right)^{1/2} \cdot \left( \sum_{k=1}^\infty |x_{k+2}|^2 \right)^{1/2}$$ and the first factor converges, but the second not neccessarily?
For the second one we have $$\sum_{k=1}^\infty |x_8+\frac{1}{k^4}x_k| \le |x_8 | + \sum_{k=1}^\infty |\frac{1}{k^4}x_k|\le \left( \sum_{k=1}^\infty |\frac{1}{k^4}| \right)^{1/2} \cdot \left( \sum_{k=1}^\infty |x_{k}|^2 \right)^{1/2}$$
and here also I don't know anything about the second factor?
Well-defined means that the operator actually maps into the space you claim it does, i.e. in your case that the image of every $x \in \ell^2$ is actually in $\ell^1$. Of course, if you show that the operator is bounded, then it is necessarily well-defined.
Now, for $T$ you actually showed that it is bounded. Indeed, you showed $$\|Tx\|_1 \leq \left(\sum\limits_{k = 1}^{\infty} \frac{1}{k^2}\right)^{1/2} \cdot \|x\|_2.$$
The second operator does not map into $\ell^1$ as you can easily see if you consider $Lx$ for $$x_j = \begin{cases} 1 & \text{for } j = 8, \\ 0 & \text{otherwise}. \end{cases}$$ Also, you made a mistake in your computation. You cannot pull the $|x_8|$ out of the sum like that.