Impulse response of a system from the input-output relation
It looks like you answered your own question but to be complete..The impulse response $H(z)=\frac{Y(z)}{X(z)}$ can be found from the difference equation which you have found from the block diagram. So taking the z-transform of both sides of the difference equation we have: $Y(z)=X(z)(\frac{2}{16}+\frac{3}{16}z^{-1}+\frac{6}{16}z^{-2}+\frac{3}{16}z^{-3}+\frac{2}{16}z^{-4})$ and so we have $H(z)=\frac{2}{16}+\frac{3}{16}z^{-1}+\frac{6}{16}z^{-2}+\frac{3}{16}z^{-3}+\frac{2}{16}z^{-4}$.

Now taking the inverse z-transform you get the impulse response to be $h[n]=\frac{2}{16}+\frac{3}{16}\delta[n-1]+\frac{6}{16}\delta[n-2]+\frac{3}{16}\delta[n-3]+\frac{2}{16}\delta[n-4]$, which has the coefficients that you listed.