1. It is a really bad idea to use a variable named l. I is hard to distinguish it from 1.
2. All thes functions using xrange are inefficient

the lcd function can be efficiently calculated by

from fractions import gcd
def lcd(a,b):
    return(a*b/gcd(a,b))

the count function is

count(l,g)=divisor_count(g/l)

where divisor_count is the number-of-divisors function

If the number n has the prime factor decomposition

$$n=p_1^{e_1}\cdot p_2^{e_2}\cdots p_k^{e_k}$$ then we have

$$ \text{divisor_count}(n)=(e_1+1)\cdot(e_2+1)\cdots(e_k+1)$$

This can be calculated in the following way:

def divisor_count(n):
    cnt=1
    i=2
    while i*i<=n:
        e=0
        while n%i==0:
            n//=i
            e+=1
        cnt*=e+1
        i+=1
    if n>1:
        cnt*=2
    return(cnt)

This divisor function runs in $$O(\sqrt{n})$$ time, the xrange implementation uses $$O(n)$$ time.

1. It is a really bad idea to use a variable named l. I is hard to distinguish it from 1.
2. All these functions using xrange are inefficient

the lcd function can be efficiently calculated by

from fractions import gcd
def lcd(a,b):
    return(a*b/gcd(a,b))

the count function is

count(l,g)=divisor_count(g/l)

where divisor_count is the number-of-divisors function

If the number n has the prime factor decomposition

$$n=p_1^{e_1}\cdot p_2^{e_2}\cdots p_k^{e_k}$$ then we have

$$ \text{divisor_count}(n)=(e_1+1)\cdot(e_2+1)\cdots(e_k+1)$$

This can be calculated in the following way:

def divisor_count(n):
    cnt=1
    i=2
    while i*i<=n:
        e=0
        while n%i==0:
            n//=i
            e+=1
        cnt*=e+1
        i+=1
    if n>1:
        cnt*=2
    return(cnt)

This divisor_count function runs in $$O(\sqrt{n})$$ time, the xrange implementation uses $$O(n)$$ time.