Step 1: Recall the reduction relation.
With $I_n=\int_0^{\pi/4}\tan^n x\,dx$, consider $I_n+I_{n+2}=\int_0^{\pi/4}\tan^n x\,(1+\tan^2 x)\,dx=\int_0^{\pi/4}\tan^n x\sec^2 x\,dx$.
Step 2: Evaluate that combination.
Let $t=\tan x$, $dt=\sec^2 x\,dx$; as $x$ goes $0\to\frac{\pi}{4}$, $t$ goes $0\to 1$. So $I_n+I_{n+2}=\int_0^1 t^n\,dt=\frac{1}{n+1}$.
Step 3: Rewrite each given term.
Thus $\frac{1}{I_2+I_4}=3$, $\frac{1}{I_3+I_5}=4$, $\frac{1}{I_4+I_6}=5$ (using $n=2,3,4$ and $\frac{1}{n+1}$).
Step 4: Add them.
The sum is $3+4+5=12$.
Step 5: Express $12$ back in the same form.
We need $\frac{1}{I_m+I_{m+2}}=12$, i.e. $\frac{1}{m+1}=\frac{1}{12}$, so $m+1=12$ and $m=11$.
Step 6: Identify the option.
Hence the value equals $\frac{1}{I_{11}+I_{13}}$, which is option (4). \[ \boxed{\dfrac{1}{I_{11}+I_{13}}} \]