Let $$f(x) = \begin{cases} \frac{\sin(x-|x|)}{x-|x|}, & x \in (-2, -1) \\ \max(2x, 3[|x|]), & |x| < 1 \\ 1, & \text{otherwise} \end{cases}$$
where $$[t]$$ denotes greatest integer $$\leq t$$. If $$m$$ is the number of points where $$f$$ is not continuous and $$n$$ is the number of points where $$f$$ is not differentiable, the ordered pair $$(m, n)$$ is:

Given,

$$f(x)= \begin{cases} \dfrac{\sin(x-|x|)}{x-|x|}, & x\in(-2,-1)\\ \max(2x,3[|x|]), &|x|<1\\ 1, & \text{otherwise} \end{cases}$$

First simplify the function in different intervals.

For $$x\in(-2,-1),$$

$$|x|=-x$$

Hence,

$$x-|x|=x-(-x)=2x$$

Therefore,

$$f(x)=\frac{\sin2x}{2x},\qquad x\in(-2,-1)$$

Now for $$|x|<1,$$

$$0\le |x|<1$$

Hence,

$$[|x|]=0$$

Therefore,

$$3[|x|]=0$$

So,

$$f(x)=\max(2x,0)$$

Thus,

$$f(x)=\begin{cases}0, & -1<x<0\\ 2x, & 0\le x<1 \end{cases}$$

Now check continuity at possible critical points.

At $$x=-2,$$

$$f(-2)=1$$

Also,

$$f(-2^+)=\lim_{x\to-2^+}\frac{\sin2x}{2x}=\frac{\sin(-4)}{-4}\neq1$$

Hence, $$f$$ is discontinuous at $$x=-2$$.

At $$x=-1,$$

$$f(-1)=1$$

Also,

$$f(-1^-)=\frac{\sin(-2)}{-2},\qquad f(-1^+)=0$$

Since left and right limits are unequal, $$f$$ is discontinuous at $$x=-1$$.

At $$x=0,$$

$$f(0^-)=0,\qquad f(0^+)=0,\qquad f(0)=0$$

Hence, $$f$$ is continuous at $$x=0$$.

At $$x=1,$$

$$f(1^-)=2,\qquad f(1)=1$$

Hence, $$f$$ is discontinuous at $$x=1$$.

Therefore, the number of discontinuity points is

$$m=3$$

Now check differentiability.

At $$x=-2,-1,1,$$ the function is discontinuous, so it is not differentiable.

At $$x=0,$$

$$f(x)=0\quad(-1<x<0)$$

and

$$f(x)=2x\quad(0<x<1)$$

Hence,

$$f'_-(0)=0,\qquad f'_+(0)=2$$

Since LHD and RHD are unequal, $$f$$ is not differentiable at $$x=0$$.

Therefore,

$$n=4$$

Hence, the ordered pair is

$$\boxed{(3,4)}$$.