Calculating impulse response of system
Working in z-domain makes the problem easy 

$$ H_{1}(z) = 1 + \frac{1}{2}z^{-1} = \frac{2z+1}{2z}$$
$$ H_{2}(z) = 1 - \frac{1}{2}z^{-1} = \frac{2z-1}{2z}$$
$$ H_{3}(z) = -1$$
$$ H_{4}(z) = \frac{-z}{z - \frac{1}{2}} = \frac{-2z}{2z-1}$$

$$ H_{3}(z) + H_{4}(z) = \frac{-4z + 1}{2z + 1} $$ 
$$ H_{2}(H_{3} + H_{4}) = (\frac{2z-1}{2z})(\frac{1-4z}{2z-1}) = \frac{1-4z}{2z}$$

$$ H(z) = H_{1} || H_{2}(H_{3} + H_{4}) = \frac{2z+1}{2z} + \frac{1-4z}{2z} =\frac{1-z}{z} $$
$$ H(z) =  z^{-1} - 1 $$
$$ \implies h[n] = -\delta[n] + \delta[n-1] $$
Hence, $a = -1, b = 1, c = 0$