Let $f(x)=\left\{\begin{array}{rr}x^3-x^2+10 x-7, & x \leq 1 \\ -2 x+\log _2\left(b^2-4\right), & x>1\end{array}\right.$ The set of values of $b$ for which $f(x)$ has greatest value at $x=1$, is

$(-6,-2) \cup(2,6)$

We have,

$f^{\prime}(x)=\left\{\begin{array}{cc} 3 x^2-2 x+10 &, ~~x<1 \\ -2 & , ~~x>1 \end{array}\right.$

Clearly, $f^{\prime}(x)>0$ for $x<1$ and $f^{\prime}(x)<0$ for $x>1$.

Thus, $f(x)$ is increasing for all $x<1$ and decreasing for $x>1$.

Therefore, $f(x)$ will have greatest value at $x=1$, if

$\lim\limits_{x \rightarrow 1^{+}} f(x)<f(1)=\lim\limits_{x \rightarrow 1^{-}} f(x)$

$\Rightarrow \lim\limits_{x \rightarrow 1}-2 x+\log _2\left(b^2-4\right)<1-1+10-7$

$\Rightarrow -2+\log _2\left(b^2-4\right)<3$

$\Rightarrow \log _2\left(b^2-4\right)<5$

$\Rightarrow b^2-4<2^5$ and $b^2-4>0$

$\Rightarrow b^2-36<0$ and $b^2-4>0$

$\Rightarrow -6<b<6$ and $b \in(-\infty,-2) \cup(2, \infty)$

$\Rightarrow b \in(-6,-2) \cup(2,6)$