Step 1: Recognize the integrand in Assertion (A) as an exact derivative.
Consider $h(x) = \frac{x}{\ln x}$. By the quotient rule: \[\frac{d}{dx}\!\left(\frac{x}{\ln x}\right) = \frac{(\ln x)(1) - x \cdot \frac{1}{x}}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2} = \frac{1}{\ln x} - \frac{1}{(\ln x)^2}\] This matches the integrand in Assertion (A) exactly.
Step 2: Evaluate the definite integral using this antiderivative.
\[\int_2^e\!\left(\frac{1}{\ln x} - \frac{1}{(\ln x)^2}\right)\! dx = \left[\frac{x}{\ln x}\right]_2^e = \frac{e}{\ln e} - \frac{2}{\ln 2} = e - \frac{2}{\ln 2}\] (since $\ln e = 1$).
Step 3: Convert to the form stated in Assertion (A).
Using the change of base formula $\log_2 e = \frac{1}{\ln 2}$, we get $\frac{2}{\ln 2} = 2\log_2 e$. So: \[\int_2^e\!\left(\frac{1}{\ln x} - \frac{1}{(\ln x)^2}\right)\! dx = e - 2\log_2 e\] Assertion (A) is TRUE.
Step 4: Verify Reason (R).
Reason (R) states that $\int_a^b e^x(f(x)+f^\prime(x))\,dx = e^b f(b) - e^a f(a)$. This follows from the standard rule $\int e^x(f(x)+f^\prime(x))\,dx = e^x f(x) + C$ (product rule in reverse), then applying the fundamental theorem. Reason (R) is TRUE.
Step 5: Check whether R explains A.
Substitute $t = \ln x$ in the Assertion (A) integral: $x = e^t$, $dx = e^t dt$. The integral becomes $\int_{\ln 2}^{1}\!\left(\frac{1}{t} - \frac{1}{t^2}\right)\!e^t\, dt$. With $f(t) = \frac{1}{t}$, $f^\prime(t) = -\frac{1}{t^2}$, the integrand is $e^t(f(t)+f^\prime(t))$. By (R): result $= [e^t \cdot \frac{1}{t}]_{\ln 2}^{1} = e - \frac{2}{\ln 2}$. So (R) correctly explains (A).
Step 6: State the final answer.
Both (A) and (R) are true, and (R) is the correct explanation of (A). \[\boxed{\text{(A) and (R) are true; (R) is the correct explanation of (A)}}\]