homework impulse response of system
First of all those sequences you provided must have a time index reference as well. I assume them to be causal sequences and $n=0$ to be the first sample. 

Then you can use different approaches to find the impulse reponse for example using the Z-transforms provide: $$ H(z) = \frac{Y(z)}{X(z)} $$
Then you have to convert $H(z)$ back into time to find the impulse response $h[n]$.

But for this simple problem, I would instead use a time-domain approach directly and compute the samples of $h[n]$ by the following procedure:

Given $x[n] = -\delta[n] + \delta[n-1]$ of length $N_x=2$ 

and $y[n] = -\delta[n-1] -\delta[n-2] + 2\delta[n-2]$ of length $N_y=4$

(note that the first sample $y[0]$ of $y[n]$ is zero but here we consider $y[0]$ to be in the support-domain of $y[n]$ no matter what its value is, the result won't change for this problem) 

1- Assuming an FIR $h[n]$ of length $L = N_y - N_x + 1 = 4-2+1 = 3$ samples.

2- Now propose the causal $h[n]$ to be $$h[n] = a\delta[n] + b\delta[n-1] + c\delta[n-2]$$

3- Now apply the convolution as the sums of shifted $h[n]$ rows as follows:

$$ y[n] = h[n]\star x[n] = x[0]h[n] + x[1]h[n-1] $$
$ $

$~~~~~~~~~~~~~~~ -a ~~ -b ~~ -c$

$~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~a ~~~~~~~b ~~~~~c$

$ + ~~~ ------------$

$y[n]=~ -a ~~(a-b) ~~ (b-c) ~~c$

$y[n]=~ -0 ~~~~(-1) ~~~~ (-1) ~~~~~2$

from which you can easily solve $a,b,c$ to be

$c = 2$ , $b-c=-1 =&gt; b = 1$ and $a=0$, therefore the impulse reponse is found to be $$ h[n] = 0\delta[n] + 1\delta[n-1] + 2\delta[n-2]$$