7. Let $f(x)=x-[x]-\tan x$, where $[x]$ is the greatest integer not exceeding $x$, then $\{T \mid f(x+T)=f(x), T \neq 0\}=$
Since the function $y=x-[x]$ has a period of 1, and the function $y=\tan x$ has a period of $\pi$, it is impossible for there to exist a $T \neq 0$ that satisfies $T$ being a period of $f(x)$. Therefore, $\{T \mid f(x+T)=f(x), T \neq 0\}=\varnothing$.