The equation for the rate constant ($\pu{s^{-1}}$) for Forster (or Resonance of dipole-dipole ) energy transfer at separation R is
$$ k_R= \alpha\frac{\kappa^2\phi}{\tau R^6}\int_0^{\infty} \frac{F(\nu)\epsilon(\nu)}{\nu^4} d\nu$$
where the constant $\alpha =(9000\ln(10) )/(128\pi^5n^4N)$, n is the solution refractive index and N Avogadro's number. The quantum yield of the donor is $\phi$, and its excited state lifetime (in the absence of quencher) is $\tau$, the orientation term is $\kappa$ and R the separation of donor and acceptor.
In the (overlap) integral F is the fluorescence spectrum measure in frequency (not wavelength) and the area under $F(\nu)$ is normalised to unity. The molar extinction coefficient of the acceptor is $\epsilon(\nu)$ also measured on a frequency scale normally in units $\pu{dm^3mol^{-1}cm^{-1}}$.
The rate constant is more commonly written as
$$k=\frac{1}{\tau} \left( \frac{R_0}{R} \right)^6 $$
where $R_0$ is the critical distance at which energy transfer rate constant equals the fluorescence rate constant ($1/\tau$) and is also a measure of the overlap of fluorescence from the donor and absorption by the acceptor.
This the decay rate of a molecule that is fluorescing and undergoing energy transfer is at separation R equal to $k=k_f+k_{isc}+k{ic}+k_R$ where isc and ic are intersystem crossing and internal conversion respectively.
In you questions;
(A) the bigger this is the larger $k_R$ is.
(B) and (C) there is no real cut off, $R_0$ for chlorophyll is $\approx 8$ nm so transfer can occur beyond this distance it just falls off as $1/R^6$
(D) Clearly the smaller $\phi$ is the lower the rate of transfer and in direct proportion. If the molecule has a low fluorescence yield clearly the molecule cannot transfer energy as it is directed elsewhere.
(E) If $\epsilon$ is small then the overlap integral is small so the rate $k_R$ is small.