Since, the roots of $x^{2}-2 m x+m^{2}-1=0$ lie between
$-2$ and 4 i.e., $\mathrm{b} 2-4 \mathrm{ac} \geq 0$ and $-2 < \frac{-\mathrm{b}}{2 \mathrm{a}} < 4$
$(2 m)^{2}-4\left(m^{2}-1\right) \geq 0$...(1)
and $-2 < \frac{2 \mathrm{~m}}{2} < 4$
$\Rightarrow-2 < m < 4$
From (i) $4 m^{2}-4 m^{2}+4 \geq 0$
$\Rightarrow m \in R$
and $\mathrm{f}(-2)>0$, also $\mathrm{f}(4)>0$
$4+4 m+m^{2}-1>0,16-8 m+m^{2}-1>0$
$\Rightarrow \mathrm{m}^{2}+4 \mathrm{~m}+3>0, \mathrm{~m}^{2}-8 \mathrm{~m}+15>0$
$\Rightarrow(m+1)(m+3)>0,(m-3)(m-5)>0$
$\Rightarrow-3 < \mathrm{m} < -1$ and $3 < \mathrm{m} < 5$
Thus, the inteval in which it lies is $-1 \leq \mathrm{m} \leq 5$