(7) $2 \sqrt{5}$ Hint: Let the four real roots of the quartic polynomial $f(x)$ be $a-3, a-1, a+1, a+3$, then
$$
f(x)=k(x-a-3)(x-a-1)(x-a+1)(x-a+3)(k \neq 0) \text {, }
$$

Let $x-a=t$, then
$$
f(x)=k(t-3)(t-1)(t+1)(t+3)=k\left(t^{4}-10 t^{2}+9\right)=g(t),
$$

Then
$$
g^{\prime}(t)=k\left(4 t^{3}-20 t\right)=4 k\left(t^{3}-5 t\right),
$$
The roots of $g^{\prime}(t)=0$ are $t=-\sqrt{5}, t=0, t=\sqrt{5}$, so the roots of $f^{\prime}(x)=0$ are $x=$ $a-\sqrt{5}, x=a, x=a+\sqrt{5}$. Therefore, the difference between the largest and smallest roots of $f^{\prime}(x)$ is $2 \sqrt{5}$.
(7) The four real roots of the quartic polynomial $f(x)$ form an arithmetic sequence with a common difference of 2. Then, the difference between the largest and smallest roots of $f^{\prime}(x)$ is