b) $f(n)=g(n)=2^{k-1}$ doesn't work because of parity a) let $\Delta_t f(n):=f(n+t)-f(n)$ (and $D_k(f)=\Delta_{2^{k-1}}(D_{k-1}(f))$, $D_0(f)=f$ It is well known that this operation decreases the degree of a polynomial. So if $d$ is the $\max\{Deg(f),Deg(g)\}$, there exists a sequence of pluses and minuses to put on $(f(1),g(1)),...,(f(2^d),g(2^d))$ so that the total vectorial sum is $0$

For b) take $f\equiv g$ and $f(1) = 1$, $f(n)$ to be any even integer for $n\geq 2$. For a) we proceed as follows. We solve the $1$-dimensional version of the problem and then immediately translate the argument to the $2$-dimensional one. Recall the well known (and very easy to check) fact that if $f(x)$ is a polynomial of degree $k$, then for any real number $a\neq 0$ the degree of $f(x+a) - f(x)$ is $k-1$ (here we conventionally treat the degree of the zero polynomial to be $(-1)$). Hence if $f_1(x) = f(x+1) - f(x)$, $f_2(x) = f_1(x+2) - f_1(x)$ and more generally $f_{i+1}(x) = f_i(x+2^i) - f_i(x)$, then $f_{k+1}$ is the zero polynomial, but is also a linear combination $\sum_{i=0}^{2^{k+1}-1}\varepsilon_i f(x+i)$ where $\varepsilon_i \in \{1,-1\}$. So by moving according to the $\varepsilon_i$-s we can infinitely often return to $0$ with a period of $2^{k+1}$. For the two dimensional version it now follows that if we analogously use only the $(\pm f(n), \pm g(n))$ operation (which we are allowed), then we return to $(0,0)$ infinitely often with a period of $2^{\max(\deg f, \deg g) +1}$, as desired.

Part (b) fails by taking $f(n)=g(n)=2^n$ because of unique binary representation. For part (a), we only have to use moves of the form $(+f(n),+g(n))$ and $(-f(n),-g(n))$. Create a sequence $a_1,a_2,\ldots$ as follows: set $a_1=1$, and then inductively construct the rest of the sequence by taking the already-determined terms, inverting them (so $1$ becomes $-1$ and vice versa), and concatenating the result (I think $a_i$ is based on the parity of the number of $1$s in $i-1$, but this doesn't matter). I then claim that picking moves according to the sequence in the obvious way works. Let $\Delta^k$ be a series of functions from $\mathbb{Z}[x]$ to itself which are defined recursively as $\Delta^0 \equiv \mathrm{id}$ and $\Delta^{k+1} f(n)=\Delta^k f(n)-\Delta^k f(n+2^k)$ for $k \geq 0$. It is clear that for some polynomial $P$ with degree $d$, $\Delta^k P$ has degree $k-d$ if $d \leq k$, and is identically zero if $d>k$. On the other hand, we equivalently have $\Delta^k f(n)=\sum_{i=0}^{2^k-1} a_{i+1}f(n+i)$. Therefore, if $D=\max(\deg f,\deg g\}$, Vlad will end up at the origin after $2^{D+1},2^{D+2},2^{D+3},\ldots$ moves. $\blacksquare$