Answer:
a)  x_{cm} = L / 2
, y_{cm}= L/2, b) x_{cm} = L / 2
, y_{cm}= L/2
Explanation:
The center of mass of a body is the point where all external forces are applied, it is defined by
        = 1 / M ∑
 = 1 / M ∑   = 1 /M ∫ x dm
 = 1 /M ∫ x dm
        = 1 / M ∑
 = 1 / M ∑  = 1 / M ∫ y dm
 = 1 / M ∫ y dm
where M is the total body mass
Let's calculate the center of mass of our L-shaped body, as formed by two rods one on the x axis and the other on the y axis
a) let's start with the reference zero at the left end of the horizontal rod
let's use the concept of linear density
     λ = M / L = dm / dl
since the rod is on the x axis
      dl = dx
     dm = λ dx
let's calculate
       x_{cm} = M ∫ x λ dx = λ / M ∫ x dx
       x_{cm} = λ / M x² / 2
we evaluate between the lower integration limits x = 0 and upper x = L
       x_{cm} = λ / M (L² / 2 - 0)
   we introduce the value of the density that is cosntnate
      x_{cm} = (M / L) L² / 2M
      x_{cm} = L / 2
We repeat the calculation for verilla verilla
      λ = M / L = dm / dy
      y_{cm} = 1 / M ∫ y λ dy
      y_{cm} = λ M y² / 2
       = M/L  1/M (L² - 0)
 = M/L  1/M (L² - 0)
      y_{cm}= L/2
b) we repeat the calculation for the origin the reference system is top of the vertical rod
      horizontal rod
         x_{cm} = 1 / M ∫λ x dx = λ/M   x² / 2
we evaluate between the lower limits x = 0 and the upper limit x = -L
       x_{cm} = λ / M [(-L)²/2 - 0] = (M / L) L² / 2M
       x_{cm} = L / 2
vertical rod
       y_{cm} = 1 / M ∫y dm
       y_{cm} = λ / M ∫y dy
       y_{cm} = λ / M y2 / 2
we evaluate between the integration limits x = 0 and higher x = -L
       y_{cm} = (M / L) 1 / M ((-L)²/2 -0)
       y_{cm} = L / 2