Response of cascading FIR filters
In this problem there are a cascade of 4 LTI systems whose multiplication of frequency responses (or convolution of impulse responses equivalently) provides the overall system Frequency Response (or impulse response equivalently) and the output $y[n]$ at n = 10 is being asked for an input of $x[n] = 0.9 \times n \times (u[n] - u[n-20])$

Now I would like to proceed by finding the overall impulse response $h[n]$ from the serial convolution of those four individual impulse responses as it seems simpler to find the output in time domain then in the Frequency domain:
$$ h[n] = (\delta[n] - \delta[n-1]) \star (\delta[n]+\delta[n-1]) \star
 (\delta[n] + \delta[n-2]) \star (\delta[n] + \delta[n-4])$$ 

Where $\star$ denotes convolution operation and those impulses are directly derived from the Frequency Responses $H_i(e^{j\omega})$ of the individual systems. I hope you can see the simple corresponce with each one of them.

Then convolving from left to right, we can show in 3 steps that:
$$h[n] = (\delta[n] - \delta[n-2]) \star (\delta[n] + \delta[n-2]) \star (\delta[n] + \delta[n-4])$$ 

$$h[n] = (\delta[n] - \delta[n-4]) \star (\delta[n] + \delta[n-4])$$

$$h[n] = (\delta[n] - \delta[n-8])$$

I hope also that you have no difficulty in obtainig shifted impulses as a result of convolution with a shifted impulse.

Therefore now the question becomes; given LTI impulse response $h[n] = \delta[n] - \delta[n-8]$ , compute $y[n]$ for the given input $x[n]$. It's now easy to show that: 
$$y[n] = h[n] \star x[n] = (\delta[n] - \delta[n-8])*x[n]$$ 
$$y[n] = x[n] -x[n-8]$$ 

therefore with $x[n] = 9 \times n \times(u[n]-u[n-20]$ , $y[10]$ becomes: 
$$y[10] = x[10] - x[2]$$
$$y[10] = 0.9 \times 10 \times (u[10]-u[-10]) - 0.9 \times 2 \times (u[2]-u[-18])$$
$$y[10] = 9 - 1.8 = 7.2$$

Where before the last line I have used the famous property of unit step $u[n]$.