7. (10 points) First, write down a sequence of numbers on the blackboard: $1,2,3, \cdots, 50$. Each time, erase the first 4 numbers, and write down the sum of the erased 4 numbers at the end of the sequence, to get a new sequence. Repeat this operation until fewer than 4 numbers remain on the blackboard. Answer the following:
（1）The sum of the numbers left on the blackboard at the end is $\qquad$ ;
（2）The last number written is . $\qquad$

保留源文本的换行和格式，直接输出翻译结果。
【Analysis】According to the problem, the sum of the 4 numbers erased each time is written at the back. In the end, only 49 and 50 are left, but the numbers after them are all sums of four numbers, totaling 12 numbers. Adding the two numbers 49 and 50, there are 14 numbers in total. Continuing the cycle, the first numbers to be erased this time are 49 and 50, as well as the sum of 1+2+3+4 and the sum of 5+6+7+8, and so on. In the end, only 4 groups of sums remain, which are: $25+26+27+28, 29+30+31+32$, $33+34+35+36, 37+38+39+40$. At this point, the sum of this group of numbers is one number, so the last remaining number is the sum of these 4 groups of numbers, which is: $25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40=520$. The last number written can be calculated from the total.

【Solution】According to the analysis, the sum of the 4 numbers erased each time is written at the back. In the end, only 49 and 50 are left, but the numbers after them are all sums of four numbers,

with a total of 12 numbers, plus the two numbers 49 and 50, making 14 numbers in total. Continuing the cycle, the first numbers to be erased this time are 49 and 50, as well as the sum of $1+2+3+4$ and the sum of 5+6+7+8, and so on. In the end, only 4 groups of sums remain, which are: $25+26+27+28=106, 29+30+31+32=122$, $33+34+35+36=138, 37+38+39+40=154$, and the sum of the four groups of numbers is: $106+122+138+154=520$. When only the following numbers remain on the blackboard: $41+42+43+44=170 ; 45+46+47+48=186 ; 49+50+1+2+3+4+5+6+7+8=$ 135 ;
$$
9+10+11+12+\cdots+24=264 ; 25+26+27+28+\cdots+40=520 ;
$$

That is, in order: $170, 186, 135, 264, 520$ five numbers, at this point, the last number written should be: $170+186+135+264$ $=755$.

So, in the end, only two numbers remain: 520 and 755, and the sum of the numbers left on the blackboard is: $520+755=1275$, so the answer is: $1275 ; 755$.