Problem 130

(1) There are three cases to consider for forming four-digit even numbers:
- Case 1: The unit digit is 0. In this case, we have to choose three different digits for the thousands, hundreds, and tens places from the remaining five digits (1, 2, 3, 4, 5). This can be done in $A_5^3$ ways, since order matters and repetition is not allowed.
- Case 2: The unit digit is 2. The thousands place can be selected from the other four digits (1, 3, 4, 5) in $A_4^1$ ways. Then, for the hundreds and tens places, we choose from the remaining three digits in $A_4^2$ ways. Therefore, there are $A_4^1 \times A_4^2$ four-digit even numbers ending in 2.
- Case 3: The unit digit is 4. This case is similar to Case 2. The thousands place is chosen from the four digits (1, 3, 4, 5) in $A_4^1$ ways, and the hundreds and tens places are selected from the remaining three digits in $A_4^2$ ways. Again, we have $A_4^1 \times A_4^2$ numbers.

Adding up the numbers from all three cases, we have the total count of four-digit even numbers as:
$$A_5^3 + A_4^1 \times A_4^2 + A_4^1 \times A_4^2 = 60 + 48 + 48 = 156.$$
Therefore, there are $\boxed{156}$ four-digit even numbers with no repeated digits.

(2) For forming five-digit numbers that are multiples of 5 and have no repeated digits, there are two cases:
- Case 1: The unit digit is 0. Since the number must be a multiple of 5, the tens digit is fixed as 0. The other four digits (1, 2, 3, 4, 5) can be arranged in the remaining four places in $A_5^4$ ways.
- Case 2: The unit digit is 5. Here, the thousands, hundreds, tens, and millions places can be chosen from the other four digits (0, 1, 2, 3, 4) in $A_4^1 \times A_4^3$ ways. We use $A_4^1$ for the first digit to ensure it's not 0.

Summing up the numbers from the two cases, we have the total count of five-digit numbers as:
$$A_5^4 + A_4^1 \times A_4^3 = 120 + 96 = 216.$$
Therefore, there are $\boxed{216}$ five-digit numbers that are multiples of 5 with no repeated digits.