I need to find the function c(k), knowing that
$$\sum_{k=0}^{\infty} \frac{c(k)}{k!}=1$$
$$\sum_{k=0}^{\infty} \frac{c(2k)}{(2k)!}=0$$
$$\sum_{k=0}^{\infty} \frac{c(2k+1)}{(2k+1)!}=1$$
$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k+1)}{(2k+1)!}=-1$$
$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k)}{(2k)!}=0$$
Is it possible?
On
You are looking for a function $\displaystyle f(z) = \sum_{k \ge 0} \frac{c(k)}{k!} z^k$ satisfying
$$f(1) = 1$$ $$f(-1) = -1$$ $$f(i) = -i.$$
Infinitely many functions have this property. There is a unique quadratic polynomial $p(z)$ with this property (for example by Lagrange interpolation), and for any entire function $q(z)$ the function $p(z) + (x - 1)(x + 1)(x - i) q(z)$ has this property. In fact these are all entire functions with this property.
More generally I think the theory of interpolation by entire functions is fairly well-understood, but I don't know of a good reference. If the set of $z$ at which you fix the value of $f$ has a limit point, then $f$ is unique by the identity theorem. If the set of $z$ at which you fix the value of $f$ is countable and does not have a limit point, then $f$ is non-unique by Weierstrass factorization.
Following Rotwang, define $b(k)=c(k)/k!$, then note that both the second and fifth can be satisfied as long as $\sum{b(2k)}=0$. Then the first and third are redundant and define $d(k)=b(2k+1)$. Now we have $\sum{d(k)}=1, \sum{(-1)^kd(k)}=-1$. Adding and subtracting, $\sum{d(2k)}=0, \sum{d(2k+1)}=1.$ So the final is that we must have $$\begin{align} \sum\frac{c(2k)}{k!}&=0\\ \sum\frac{c(4k+1)}{(4k+1)!}&=0\\ \sum\frac{c(4k+3)}{(4k+3)!}&=1 \end{align}$$
and any $c(k)$ that satisfies this will work.