I am aware that this question has been asked already here, however there is no accepted answer to it yet. I have no idea where to start.
we know that
$$ f(t)= \lambda \times exp (-\lambda t)$$ then
$$ F(t)= 1 - exp (-\lambda t)$$
but i am not sure how to continue
Because $T$ is continuous, $$P(X = k) = P(k \le T < k+1) = F_T(k+1) - F_T(k)\\ = e^{-k\lambda} - e^{-(k-1)\lambda},$$ for $\lambda > 0$ and $k = 0, 1, \dots,$ where you have given $F_T$ in your Question.
The process you have in mind amounts to rounding exponential data down to the next lower integer; that is, taking the 'floor' function.
For exponential rate $\lambda = 1/5,$ one has $E(T) = 5$ and $V(T) = 25.$ The following computation in R evaluates $E(X) \approx 4.517$ and $Var(X) \approx 24.92$ by summing the first 100 terms in the relevant infinite series. An exact analytic evaluation is possible.
Also shown are the first ten probabilities. [In R,
pexpis an exponential CDF.]A simulation of a million observations from the distribution of $X$ gives approximations of the above (mostly accurate to two or three significant digits).