You have the right idea, but your notation is badly confused. First, the relation $\sim$ is a relation between Cauchy sequences, not between equivalence classes of them. You can write $(x_i)\sim(y_i)$, for instance, or $[(x_i)]=[(y_i)]$, which means exactly the same thing, but you can’t write $[(x_i)]\sim[(y_i)]$: it’s meaningless.
Next, as Martin suggested, you don’t want [(x_i)]': that suggests that the Cauchy sequence $(x_i)$ is generating two equivalence classes, $[(x_i)]$ and [(x_i)]'. It would be simpler to avoid the primes altogether and just use a different letter from $x$.
For (a) you’re basically right: you want to define $-[(x_i)]$ to be $[(-x_i)]$. You’re also on the right track for (b): to show that this is well-defined, you must show that you get the same result no matter which member of the equivalence class $[(x_i)]$ you use to define $-[(x_i)]$. In other words, you must show that if $(x_i)\sim(y_i)$, then $(-x_i)\sim(-y_i)$.
To do this, translate the hypothesis, that $(x_i)\sim(y_i)$, and the desired conclusion, that $(-x_i)\sim(-y_i)$, into statements involving more basic notions by using the definition of $\sim$: you want to show that $\lim_{n\to\infty}|x_n-y_n|=0 \implies \lim_{n\to\infty}|(-x_n)-(-y_n)|=0\;.$ This should be pretty straightforward.