Linear Time-Invariant system frequency response and impulse response given input and output in time domain
A classical question on signals and systems, for which I provide you one method of solution.

Now given an LTI system, and its input $x[n]$ and output $y[n]$, the fundamental approach to determine the impulse response $h[n]$ involves the utilisation of a frequency domain approach to assert $H(e^{jw}) = Y(e^{jw})/X(e^{jw})$

In this problem however, I would choose a different approach which involves more of an analysis by inspection rather than pure mathematical operations, because of the fact that the input $x[n]$ involves a cosine term whose DTFT is includes impulses and will be problematic in the division process.

The method proceeds by drawing the DTFTs of input and output:
The input $x[n]$ involves two terms one of which is $$5 \sin(0.4 \pi n)/ \pi n$$ whose DTFT is a zero phase ideal lowpass filter of cutoff frequency $W_c = 0.4 \pi$ with a gain of 5. 

And the second term of input is $$10 \cos (0.5 \pi n)$$ whose DTFT is modeled as two impulses at $$ 5 \pi \delta(\omega + 0.5 \pi) +  5 \pi \delta(\omega - 0.5 \pi) $$ Because of linearity property of DTFT, these two DTFT are added to represent the DTFT of the input $x[n]$.

Then, look at the DTFT of the output $$y[n] = \frac {10 \sin(0.3 \pi (n-10) )}{\pi (n-10)} $$ which is again an ideal low pass filter of gain = 10 and $W_c = 0.3 \pi$ with a phase term of $e^{-10 \omega}$ hence $Y(e^{j\omega}) =10 e^{-j10 \omega}$ when $|\omega| &lt; 0.3 \pi$ and zero elsewhere.

Now we want to find a frequency response $H(e^{j\omega})$ such that when it is multiplied by the DTFT $X(e^{j\omega})$ of the input $x[n]$, it should produce the DTFT $Y(e^{j\omega})$ of the output $y[n]$.

The observation of the input and output DTFTs yields the conclusion that one such $H(e^{j \omega})$ is a linear phase ideal lowpass filter of gain = 2, $W_c = 0.3\pi$ and phase term of $e^{-j10 \omega}$, whose time domain correspondence is found by inspection as $$h[n] = \frac {2 \sin(0.3 \pi (n-10))}{\pi (n-10)}$$

The justification of this solution can be made by observing the effect of this filter on its input: it will completely supress any signal above the frequency $\omega = 0.3 \pi$ and it will multiply any signal below that frequency by 2, while adding a phase shift of -10 radians which reflects itself as a group delay of 10 samples in (n-10) term.