1. We are given $f(x)=|[x]|+\sqrt{x-[x]},$ where $[x]$ is the greatest integer function.

2. Let $[x]=n, \quad n\in \mathbb{Z}.$ Then for $x\in[n,n+1)$, $x-[x]=x-n \in [0,1),$ so $f(x)=|n|+\sqrt{x-n}.$

3. On each open interval $(n,n+1)$, the value of $[x]$ is constant, hence $f(x)=|n|+\sqrt{x-n}$ is continuous there.

   Therefore, discontinuities can occur only at integer points.

4. We need integer points in the interval $(-2,1)$. These are $-1,\;0.$

5. Check continuity at $x=-1$:

   • For $x\in[-2,-1)$, we have $[x]=-2$, so $f(x)=|-2|+\sqrt{x+2}=2+\sqrt{x+2}.$ Thus, $\lim_{x\to -1^-} f(x)=2+\sqrt{1}=3.$

   • For $x\in[-1,0)$, we have $[x]=-1$, so $f(x)=|-1|+\sqrt{x+1}=1+\sqrt{x+1}.$ Thus, $\lim_{x\to -1^+} f(x)=1+0=1.$

   Since $\lim_{x\to -1^-} f(x)\ne \lim_{x\to -1^+} f(x),$ $f$ is discontinuous at $x=-1$.

6. Check continuity at $x=0$:

   • For $x\in[-1,0)$, we have $[x]=-1$, so $f(x)=1+\sqrt{x+1}.$ Hence, $\lim_{x\to 0^-} f(x)=1+1=2.$

   • For $x\in[0,1)$, we have $[x]=0$, so $f(x)=0+\sqrt{x}=\sqrt{x}.$ Hence, $\lim_{x\to 0^+} f(x)=0.$

   Since $\lim_{x\to 0^-} f(x)\ne \lim_{x\to 0^+} f(x),$ $f$ is discontinuous at $x=0$.

7. There are no other integers inside $(-2,1)$, so no other discontinuity points in this interval.

Therefore, the number of discontinuity points in $(-2,1)$ is $2.$