a. Let $F_k(n)$ be the collection of sequences/vectors $x$ corresponding to $f_k(n)$. Suppose $x = (x_1, ..., x_n) \in F_1(n)$. Then at least one of the entries $x_i \ne 0$. For the first such nonzero value, change its sign and call the resulting sequence $\alpha(x)$. Since $\sum x_i \equiv 1 \pmod 4$, we have $\sum \alpha(x) \equiv 3 \pmod 4$. The map $\alpha$ is also an involution. Thus
$$f_1(n) = f_3(n). \qquad (5)$$

b. Applying the same map $\alpha$ to the set $F_0(n) - \bar{0}$ (where $\bar{0}$ is the sequence of all zero's) gives a bijection between $F_0(n) - \bar{0}$ and $F_2(n)$. Thus
$$f_0(n) - 1 = f_2(n). \quad (6)$$
Since these sum residue classes partition all such sequences we have
$$f_0(n) + f_1(n) + f_2(n) + f_3(n) = 3^n. \quad (7)$$
One more equation is required. Let $A_n = F_0(n) \cup F_2(n)$, the set of all sequences that have an even sum, and let $B_n = F_1(n) \cup F_3(n)$, the set of all sequences that have an odd sum.
For a sequence $x = (x_1, \dots, x_n)$, let $\beta(x)$ be the sequence obtained by finding the first element of the sequence $x$ that is not 1, and changing it with the third entry (i.e. $-1 \mapsto 0$ or $0 \mapsto -1$). The operation $\beta(x)$ is well defined for $x \neq (1, \dots, 1)$. It also changes the parity of the sum.
Thus $\beta: A_n - \bar{1} \mapsto B_n - \bar{1}$ is a bijection. The parity of $n$ tells us which of $A_n$ and $B_n$ the vector $\bar{1}$ is in, hence
$$f_0(n) + f_2(n) = f_1(n) + f_3(n) + (-1)^n. \quad (8)$$
Combining (1), (2), (3) and (4) give
$$f_0(n) = \frac{3^n + 2 + (-1)^n}{4}. \quad (9)$$