Step 1: Look at the integral.
We must find $\int_2^e\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right]dx$ and then match it to $a+\frac{b}{\log 2}$.
Step 2: Choose a substitution.
Let $\log x=t$. Then $x=e^t$ and $dx=e^t\,dt$. This turns the messy logs into a clean form.
Step 3: Change the limits.
When $x=2$, $t=\log 2$. When $x=e$, $t=\log e=1$. So the new limits are from $\log 2$ to $1$.
Step 4: Rewrite the integral.
\[ I=\int_{\log 2}^{1}\left(\frac{1}{t}-\frac{1}{t^2}\right)e^t\,dt. \]
Step 5: Spot the special pattern.
This matches $\int e^t\big[f(t)+f'(t)\big]dt=e^t f(t)+c$ with $f(t)=\frac{1}{t}$ and $f'(t)=-\frac{1}{t^2}$. So the antiderivative is $\frac{e^t}{t}$.
Step 6: Apply the limits.
\[ I=\left[\frac{e^t}{t}\right]_{\log 2}^{1}=\frac{e^1}{1}-\frac{e^{\log 2}}{\log 2}. \] Since $e^{\log 2}=2$, \[ I=e-\frac{2}{\log 2}. \]
Step 7: Compare and read off $a,b$.
Matching $a+\frac{b}{\log 2}=e-\frac{2}{\log 2}$ gives $a=e$ and $b=-2$, which is option (2).
\[ \boxed{a=e,\ b=-2} \]