Call the digits:

    a b c

    d e f

    g h i

For each of the 6 3-digit numbers, the first digit cannot be zero, the sum of the 3 digits must be divisible by 3, and furthermore, the final digits [c,f,g,h,i] must each be even.  If those conditions are met, all 6 will be divisible by 6.

The digits [a,b,d,e] must each be odd.

To be divisible by 5, the final digit must be 0 or 5, but 5 is ruled out, since final digits must be even.

The sums of (abc), (def), and (ghi) must each be 0 mod 3.  So the sum of all 9 digits is 0 mod 3.  Given that all 9 digits are distinct, the missing digit must be 0 mod 3.  But the 5 cells in the right column and bottom row must all be even, so the missing digit must be odd and can only be either 3 or 9.

So far, we know that exactly one of [c,f,g,h,i] must be zero.  In 4 of these cases, 1 number is divisible by 5.  But if i=0 then 2 numbers are divisible by 5.

Analytic Method:  assume all ways of placing the 0 into those 5 positions is equally likely (probably not a valid assumption).

Then the number of 3-digit numbers divisible by 5 is the average of [1,1,1,1,2] = 1.2   <--  but this is wrong

    

Computer Method: confirm or refute with brute force program.

The computer program shows that the above assumption was indeed incorrect.

The brute force program shows that the 0 is never in the bottom right corner (i=0 never occurs) and so in each case, there is exactly one number divisible by 5.

Program output showing total, singles, doubles:

64 64 0

from itertools import permutations

ct_all = 0 

ct_one = 0 

ct_two = 0 

evens = '02468'

for p in permutations('0123456789',9):    

    a = int(p[0])

    b = int(p[1])

    c = int(p[2])

    if c % 2 != 0:

        continue

    if (a+b+c) % 3 != 0:

        continue

    d = int(p[3])

    e = int(p[4])

    f = int(p[5])

    if f % 2 != 0:

        continue

    if (d+e+f) % 3 != 0:

        continue

    g = int(p[6])

    if g % 2 != 0:

        continue

    h = int(p[7])

    if h % 2 != 0:

        continue

    i = int(p[8] )

    if i % 2 != 0:

        continue

    if (g+h+i) % 3 != 0:

        continue

    if (a+d+g) % 3 != 0:

        continue

    if (b+e+h) % 3 != 0:

        continue

    if (c+f+i) % 3 != 0:

        continue

    ct_all += 1

    if c==0 or f==0 or g==0 or h==0:

        ct_one += 1

    if i==0:

        ct_two += 1

        print(a,b,c,d,e,f,g,h,i)

print(ct_all, ct_one, ct_two)