Introduction:
In this problem, we are given that k is a proper subgroup of H, and H is a proper subgroup of G. We are also given the orders of k and G, and we need to determine the possible orders of H.

Definitions:
Before we proceed with the solution, let's define some terms:
- A proper subgroup is a subgroup that is not equal to the group itself. In other words, it is a strict subset of the group.
- The order of a subgroup is the number of elements in the subgroup.

Solution:
Since k is a proper subgroup of H, we know that |k| < |h|.="" similarly,="" since="" h="" is="" a="" proper="" subgroup="" of="" g,="" we="" know="" that="" |h|="" />< |g|.="" these="" inequalities="" give="" us="" the="" />

|k| < |h|="" />< />

We are given that |k| = 42 and |G| = 420. Let's consider the possible values of |H| based on these inequalities.

Divisors of 42:
The order of k is 42, which means it has 42 elements. We know that |k| < |h|,="" so="" the="" possible="" values="" of="" |h|="" should="" be="" one="" of="" the="" divisors="" of="" 42.="" the="" divisors="" of="" 42="" />

1, 2, 3, 6, 7, 14, 21, 42

Divisors of 420:
The order of G is 420, which means it has 420 elements. We know that |H| < |g|,="" so="" the="" possible="" values="" of="" |h|="" should="" be="" one="" of="" the="" divisors="" of="" 420.="" the="" divisors="" of="" 420="" />

1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210, 420

Possible Orders of H:
Based on the above information, the possible orders of H can be obtained by finding the common divisors of 42 and 420. These common divisors are:

1, 2, 3, 6, 7, 14, 21, 42

Therefore, the possible orders of H are 1, 2, 3, 6, 7, 14, 21, and 42.

Conclusion:
In this problem, we found the possible orders of H given that k is a proper subgroup of H, and H is a proper subgroup of G. The possible orders of H are 1, 2, 3, 6, 7, 14, 21, and 42, which are the common divisors of the orders of k and G.