Step 1: Observe the integrand in Assertion (A).
We have
\[
\int_{2}^{e}\left(\frac{1}{\log_e x}-\frac{1}{(\log_e x)^2}\right)\,dx
\]
Since
\[
\log_e x=\ln x,
\]
the integral becomes
\[
\int_{2}^{e}\left(\frac{1}{\ln x}-\frac{1}{(\ln x)^2}\right)\,dx
\]
Step 2: Identify the derivative form.
Consider
\[
\frac{x}{\ln x}
\]
Differentiating,
\[
\frac{d}{dx}\left(\frac{x}{\ln x}\right)
=
\frac{(\ln x)(1)-x\cdot \frac{1}{x}}{(\ln x)^2}
\]
\[
=
\frac{\ln x-1}{(\ln x)^2}
\]
\[
=
\frac{1}{\ln x}-\frac{1}{(\ln x)^2}
\]
So,
\[
\frac{1}{\ln x}-\frac{1}{(\ln x)^2}
=
\frac{d}{dx}\left(\frac{x}{\ln x}\right)
\]
Step 3: Evaluate the definite integral.
Thus,
\[
\int_{2}^{e}\left(\frac{1}{\ln x}-\frac{1}{(\ln x)^2}\right)\,dx
=
\left[\frac{x}{\ln x}\right]_{2}^{e}
\]
\[
=
\frac{e}{\ln e}-\frac{2}{\ln 2}
\]
Since
\[
\ln e=1,
\]
we get
\[
=
e-\frac{2}{\ln 2}
\]
Also,
\[
\log_2 e=\frac{1}{\ln 2}
\]
Therefore,
\[
e-\frac{2}{\ln 2}
=
e-2\log_2 e
\]
Hence, Assertion (A) is true.
Step 4: Verify Reason (R).
Consider
\[
\frac{d}{dx}\left(e^x f(x)\right)
\]
Using product rule,
\[
\frac{d}{dx}\left(e^x f(x)\right)
=
e^x f(x)+e^x f'(x)
\]
\[
=
e^x(f(x)+f'(x))
\]
Therefore,
\[
\int_a^b e^x(f(x)+f'(x))\,dx
=
\left[e^x f(x)\right]_a^b
\]
\[
=
e^bf(b)-e^af(a)
\]
Hence, Reason (R) is also true.
Step 5: Check whether Reason explains Assertion.
The assertion follows from the same product-rule idea, because the integrand is the derivative of
\[
\frac{x}{\ln x}
\]
Thus, Reason (R) gives the correct method behind Assertion (A).
Step 6: Final conclusion.
Therefore,
\[
\boxed{\text{(A) and (R) are true, and (R) is the correct explanation to (A).}}
\]