The existence of bounded linear functional on a complex Hilbert space

In both cases a linear functional with the given values on the set of the ${e_n}$ will (due to linearity) necessarily satisfy $$f\left(\sum_{i=1}^N \lambda_i e_i\right) = \sum_{i=1}^N \lambda_i f(e_i)$$ and if $f$ is bounded (which implies continuity) the only way to define it for infinite sums looks the same with $N$ replaced by $\infty$.

In the second example, if you want to show that this cannot be a bounded functional, it is sufficient to find an example for which the sums $f\left(\sum_{i=1}^N \lambda_i e_i\right) $ cannot be bounded - for that you were already on track with the example you mention.

In case $f(e_k) = \frac{1}{k}$ note that $$ \left|\sum_{n=1}^N \lambda_n f(e_n) \right|= \left|\sum_{n=1}^N \frac{\lambda_n}{n} \right| \le \sum_{n=1}^N \left|\frac{\lambda_n}{n}\right| \le \sqrt{\sum_{n=1}^N |\lambda_n|^2 }\sqrt{\sum_{n=1}^N\frac{1}{n^2}} $$ This should be more than enough of a hint to complete the exercise.