I'm not sure.I think the solution is a)480 b)1

divide in "blocks" the written digits: the first is 1995, the second 11999955... clearly at the $kth$ block the sum of its digits is $24k$. So the sum of all digits written from the first to the $kth$ block is $24 \left( \frac{k(k+1)}{2} \right)$. Let it be equal to 2880, and obtain $k=15$. So we have to write 15 blocks of digits. In the first block we write 4 digits, in the second 8, in the third 12, in the $kth$ block $4k$ digits. So the number of digits written from first to $kth$ block is $4 \left( \frac{k(k+1)}{2} \right)$. Substitute $k=15$ to obtain 480 digits

the greatest integer $\leq 1995$ that can be written in the form $4 \left( \frac{k(k+1)}{2} \right)$ is 1984 with $k=31$. So the 1984th digit is a 5. Note that in the first block the first digit is 1, in the second the first 2 digits are 1, in the $kth$ block the first $k$ digits are 1. Since $1995-1984=15<31$, the 1995th digit is 1

1 Let the equation $2x^2+2x$ represent the amount of digits per iteration. $\displaystyle\sum_{n=1}^{y}24n=2880$ Factor: $24(\frac{(y)(y+1)}{2})=2880$ $y=15$ Plug back- $2(15)^2+2(15)=\boxed{480}$.

2 Let $2x^2+2x=1995$ $x=\frac{\sqrt{3991}-1}{2}\approx 31.087$ Last digit is $\boxed{1}$ since $.087<.25$

$1$