What am I counting wrong? EDIT: I made a mistake in the beginning, the second condition has changed. Sorry for this.
I'm asked to count the number of sets of 4 elements that satisfy the two following conditions:
1) Each element of the set is a two-digit number (from 10 to 99).
2) There are no repeated digits in the set.
What I've done is separate the sets in two disjoint cases. The ones that have an element with the digit 0 and the ones that don't.
The sets from the first case have three elements that don't have a $0$ and one that does. For the first element of the three mentioned, there are $C(7,2)$ ways of choosing the digits it will have and then two possible numbers they can form, so $2C(7,2)$ posible elements. Applying the same reasoning to the rest there are $2C(5,2)$ and $2C(3,2)$ possible elements. Order doesn't matter in a set, so I divide by $4!$. 
This means that the number of sets from the first case is: $C(9,7)C(7,2)C(5,2)C(3,2)\cdot2^3\cdot1/4!$
Then I counted the number of sets from the second case using a similar argument and add them together:
$C(9,7)C(7,2)C(5,2)C(3,2)2^3\cdot1/4! + C(9,8)C(8,2)C(6,2)C(4,2)\cdot2^4\cdot1/4!=2^3\cdot3^4\cdot5\cdot7$
The answer is supposed to be twice that. What is it I'm I doing wrong?
 A: Case $1$ - digit $0$ appears:$$\frac{4\cdot\binom97\cdot7!}{4!}$$
Case $2$ - digit $0$ does not appear: $$\frac{\binom98\cdot8!}{4!}$$
Total count:$$\frac{4\cdot\binom97\cdot7!+\binom98\cdot8!}{4!}=\frac{3\cdot9!}{4!}=45360=2^4\cdot3^4\cdot5\cdot7$$
As for the error in your counting, you seem to have mixed your cases.
For case $1$ you posted: $$C(9,7)C(7,2)C(5,2)C(3,2)\cdot2^3\cdot1/4!$$
And for case $2$ you have $$C(9,8)C(8,2)C(6,2)C(4,2)\cdot2^4\cdot1/4!$$
But case $1$ should have been:  

The first element contains the digit $0$ and there are $C(9,1)$ choices for the other digit. Then $2C(8,2)$ for the second element, $2C(6,2)$ for the third and $2C(4,2)$ for the fourth. The element with $0$ is fixed in the first postion and the order of the other three does not matter, so we divide by $3!$

The number of sets in the first case is: $$\frac{C(9,1)\cdot2C(8,2)\cdot2C(6,2)\cdot2C(4,2)}{3!}\tag{1}$$
Similarly, case 2 should be:

There are $2C(9,2)$ choices for the first element, $2C(7,2)$ for the second, $2C(5,2)$ for the third and $2C(3,2)$ for the fourth. The order does not matter, so we divide by $4!$

The number of sets in the second case is: $$\frac{2C(9,2)\cdot2C(7,2)\cdot2C(5,2)\cdot2C(3,2)}{4!}\tag{2}$$
Adding these gives the correct answer of $2^4\cdot3^4\cdot5\cdot7=45360$
I hope that clears it up for you.
Now, to relate $(1)$ and $(2)$ with the solution I posted, I'll rewrite them using only factorials:
$$\require{cancel}\frac{\frac{9!}{1!\cdot\cancel{8!}}\cdot\bcancel2\cdot\frac{\cancel{8!}}{\bcancel{2!}\cdot\cancel{6!}}\cdot\bcancel2\cdot\frac{\cancel{6!}}{\bcancel{2!}\cdot\cancel{4!}}\cdot\bcancel2\cdot\frac{\cancel{4!}}{\bcancel{2!}\cdot2!}}{3!}=\frac{\frac{9!}{2!}}{3!}=\frac{4\cdot\frac{9!}{2!}}{4!}=\frac{2\cdot9!}{4!}\tag{1}$$
$$\frac{\bcancel2\cdot\frac{9!}{\bcancel{2!}\cdot\cancel{7!}}\cdot\bcancel2\cdot\frac{\cancel{7!}}{\bcancel{2!}\cdot\cancel{5!}}\cdot\bcancel2\cdot\frac{\cancel{5!}}{\bcancel{2!}\cdot\cancel{3!}}\cdot\bcancel2\cdot\frac{\cancel{3!}}{\bcancel{2!}\cdot1!}}{3!}=\frac{9!}{4!}\tag{2}$$
$$\frac{2\cdot9!}{4!}+\frac{9!}{4!}=\frac{3\cdot9!}{4!}\tag{1+2}$$
A: You have made two or three errors in interpreting the question (assuming the question is exactly as you stated:
(1) In your sets involving zero, you seem to assume that there are precisely four elements in each set.  Thus, your counting would exclude sets like $\{10, 32\}$.
(2) In your sets involving no zero, you again assume precisely four elements in each set. (Are you sure the problem did not specify only sets with four elements?)  (you have made this correction now)
(3) Your stated conditions do not rule out sets like $\{10, 23, 45, 88\}$ since the digit $8$ does not appear in any element other than the $88$ itself. (No two elements share the same digit.)
Finally, I believe the answer to the problem, even restricted to four elements in each set, is not $2\cdot (2^2\cdot 3^4 \cdot 5 \cdot 7)$ in any case. Restricting to require no digit used twice (which is a stronger restriction than you state) we get
$$
9\cdot \binom{8}{2} \cdot \binom62 \cdot \binom42 \cdot \frac86
+ \binom92 \cdot \binom72 \cdot \binom52 \cdot \binom32 \cdot \frac{16}{34} = 60480 = 2^6\cdot 3^3\cdot 5 \cdot 7
$$
