Let's analyze the equation $ |x+1||x+3|-4|x+2|+5=0 $ based on different intervals of $ x $.

(I) If $ x < -3 $ :

$ \begin{aligned} & |x+1| = -(x+1), \\ & |x+3| = -(x+3), \\ & |x+2| = -(x+2), \\ & -(x+1)*(-(x+3)) - 4(-(x+2)) + 5 = 0 \\ & x^2 + 4x + 3 + 4x + 8 + 5 = 0 \\ & x^2 + 8x + 16 = 0 \\ & \Rightarrow x = -4 \quad (\text{one solution}) \end{aligned} $

(II) If $ -3 \leq x < -2 $ :

$ \begin{aligned} & |x+1| = -(x+1), \\ & |x+3| = -(x+3), \\ & |x+2| = -(x+2), \\ & -(x+1)*(-(x+3)) - 4(-(x+2)) + 5 = 0 \\ & x^2 - 10 = 0 \\ & \Rightarrow x = \pm \sqrt{10} \\ & \text{(Do not satisfy } -3 \leq x < -2) \end{aligned} $

(III) If $ -2 \leq x < -1 $ :

$ \begin{aligned} & |x+1|=-(x+1), \\ & |x+3|=x+3, \\ & |x+2|=-(x+2), \\ & -(x+1)(x+3)-4(-(x+2))+5=0 \\ & -x^2-4x-3-4x-8+5=0 \\ & -x^2-8x-6=0 \\ & \Rightarrow x^2+8x+6=0 \\ & x=\frac{-8 \pm 2\sqrt{10}}{2} = -4 \pm \sqrt{10} \end{aligned} $

(IV) If $ x \geq -1 $ :

$ \begin{aligned} & |x+1|=x+1, \\ & |x+3|=x+3, \\ & |x+2|=x+2, \\ & (x+1)(x+3)-4(x+2)+5=0 \\ & x^2+4x+3-4x-8+5=0 \\ & x^2=0 \\ & \Rightarrow x=0 \quad (\text{one solution}) \end{aligned} $

$\Rightarrow$ The number of distinct real roots are two : $ x = -4 $ and $ x = 0 $.