Solution:

By plugging $m=n=0$ into (b) we easily get $f(0)=0$. For any $u \in \mathbb{Z}$, we have
$$\begin{aligned} f(u+1)^2 - f(u-1)^2 & = f(2u) f(2) \\ f(u+1)^2 - f(u)^2 & = f(2u+1) f(1) = f(2u+1) \\ f(u)^2 - f(u-1)^2 & = f(2u-1) f(1) = f(2u-1) \end{aligned}$$
whence
$$f(2u) f(2) = f(2u+1) + f(2u-1) .$$
We would like to conclude that
$$f(n+1) + f(n-1) = f(2) f(n)$$
for all $n \in \mathbb{Z}$. This is indubitable if $n$ is even; otherwise we may use (c) and the fact that $n+2013$ is even.
For any given value of $t = f(2)$, there is a unique function $f$ satisfying the recursive definition
$$f(1) = 1, \quad f(2) = t, \quad f(n+1) + f(n-1) = t f(n) .$$
If $t \neq \pm 2$, this solution is given by
$$f(n) = \frac{\lambda_1^n - \lambda_2^n}{\lambda_1 - \lambda_2} \quad \text{ where } \lambda_{1,2} \in \mathbb{C} \text{ are the roots of } \lambda^2 - t \lambda + 1 = 0$$
Those familiar with the theory of linear recurrences will know a heuristic derivation of this formula. For our purposes it suffices to note that this function $f$ does indeed satisfy definition (3) and in fact the condition (b) as well; thus the problem is to find out how many values of $t$ cause condition (c) to hold.
If $t = \pm 2$, the solution (4) is invalid due to the fact that $\lambda_1 = \lambda_2$. In these cases the corresponding functions $f$ satisfying (3) are $f(n) = n$ and $f(n) = (-1)^{n+1} n$, both of which fail condition (c) and hence can be discarded.
From the condition $f(2013) = f(0) = 0$, we derive that $\lambda_1^{2013} = \lambda_2^{2013} = \lambda_1^{-2013}$, so $\lambda_1^{4026} = 1$. We must have $\lambda^{2013} = 1$ or $\lambda^{2013} = -1$. If the latter holds, then from $f(2014) = f(1)$ we get
$$\begin{aligned} \lambda_1^{2014} - \lambda_2^{2014} & = \lambda_1 - \lambda_2 \\ \lambda_1 \cdot \lambda_1^{2013} - \lambda_2 \cdot \lambda_2^{2013} & = \lambda_1 - \lambda_2 \\ -\lambda_1 + \lambda_2 & = \lambda_1 - \lambda_2 \\ \lambda_1 & = \lambda_2, \end{aligned}$$
a contradiction. So $\lambda_1$, and hence its reciprocal $\lambda_2$, are 2013th roots of unity, a condition that is clearly sufficient to imply (c).
The trivial root $\lambda_1 = \lambda_2 = 1$ must be discarded. The remaining roots come in 1006 conjugate pairs yielding 1006 distinct real values of $t$. We conclude that there are 1006 such functions $f$.