Each series is based on a common series, for example: S=1+11+111+1111+... So if we work out Sn for two of the series we only need to multiply by 5 and 7 respectively to get the sum to n terms of each.
We can see that S consists of the sum of several GPs: 1+10+100+... is a GP with sum (10^n-1)/9, where n starts at 1. The next series is (10^(n-1)-1)/9. If, for example, we take n=2, we get (10^2-1)/9+(10-1)/9=11+1. For n=5, we get (10^5-1)/9+(10^4-1)/9+(10^3-1)/9+(10^2-1)/9+(10-1)/9=11111+1111+111+11+1.
What we have here is 10^5/9+10^4/9+...+10/9-5/9. This is a GP with a=10/9 and r=10 less 5/9. The sum of the GP by itself is (10/9)(10^5-1)/9=10(10^5-1)/81=11110/9. Subtract 5/9 from this: 12345. The general expression is 10(10^n-1)/81-n/9. For the first and third series then, the sum to n terms is 50(10^n-1)/81-5n/9 and 70(10^n-1)/81-7n/9.
The middle series is slightly different in that r=0.1 instead of 10.
S=3(0.1+0.11+...)=3((0.1)+(0.1+0.01)+(0.1+0.01+0.001)+...). Here a=0.1, r=0.1 for each sub-series.
As in the earlier exercise, take n=5: the sum of the main series is 0.1((1-0.1)/0.9+0.1(1-0.1^2)/0.9+0.1(1-0.1^3)0.9+0.1(1-0.1^4)/0.9+0.1(1-0.1^5)/0.9=0.1((1-0.1)/0.9+...+(1-0.1^5)/0.9)=0.1(5/0.9-(0.1^5)/0.9-(0.1^4)/0.9-...-(0.1)/0.9)=0.1(5/0.9-S'), where S' is the GP with a=0.1/0.9=1/9 and r=0.1: 0.1/0.9+0.1^2/0.9+0.1^3/0.9+0.1^3/0.9+0.1^4/0.9=(1/9)(1+0.1+...+0.1^(5-1))
or for n: (1/9)(1+0.1+...+0.1^(n-1))=(1/9)(1-0.1^n)/(1-0.1)=(1-0.1^n)/8.1.
So the complete series is 0.1(n/0.9-(1-0.1^n)/8.1). S is therefore 0.3(n/0.9-(1-0.1^n)/8.1).
(When n is large, this approximates to n/3-1/27 or (9n-1)/27.)
[Apologies for the length of time to produce the complete solution.]