Expected profit of dice game

Is the number of desired events equal to 3 ?

No, its not. You are counting the number of dice that shows a given number, after throwing three dice. The number of events is four: zero, one, two or three dice shows the chosen number. And the choosing of the number is not a probability event at all, what is probabilistic is the number of dice with the chosen number.

Do we multiply each of these outcomes with the corresponding probabilities and sum them up?

Yes, its exactly that. There is a theorem than says that

$$ \mathrm{E}[h(Y)]=\int_{\mathbb{R}}h(y)P(dy) $$

In the case of $Y$ discrete then the above integral can be written as the sum

$$ \mathrm{E}[h(Y)]=\sum_{k\in \mathbb{N}}h(k)\Pr [Y=k] $$

Now, if $Y$ is the probability distribution that count the number of "good numbers" after throwing three dice, then its easily seen that $X=h(Y)$ for some function $h$.

(c) We get the formula of (b) and instead of 3 we write c and the result must be 0 ?

Yes, that is. Now you must choose $c$ such that $\mathrm{E}[X]=0$.