Yes, you are given that \(\displaystyle P(1)= \frac{c}{3}\), \(\displaystyle P(2)= \frac{c}{3^2}=\frac{c}{9}\), etc. The condition on c is that the "total" probability is \(\displaystyle \frac{c}{3}+ \frac{c}{9}+ \cdot\cdot\cdot+ \frac{c}{3^n}+ \cdot\cdot\cdot= c(\frac{1}{3}+ \frac{1}{3^3}+ \cdot\cdot\cdot+ \frac{1}{3^n}+ \cdot\cdot\cdot)\). That is a geometric series except that it begins at n= 1 rather than n= 0. The sum of a geometric sequence with initial term c and "common ratio" \(\displaystyle \frac{1}{3}\) is \(\displaystyle \frac{c}{1-\frac{1}{3}}= \frac{3c}{2}\). Since this series is missing the first term, c, its sum is \(\displaystyle \frac{3c}{2}- c= \frac{c}{2}\). That must be 1 so, yes, \(\displaystyle c= 2\).
Now, the probability that \(\displaystyle X\ge k\) is simply that minus the probability that \(\displaystyle X< k\). The probability that \(\displaystyle X< k\) is the sum \(\displaystyle \frac{2}{3}+ \frac{2}{9}+ \cdot\cdot\cdot+ \frac{2}{3^{k-1}}\) which is a finite geometric sum. The sum of such a finite geometric sum, with initial term a and common ratio r, from 0 to n, is \(\displaystyle \frac{a(1- r^{n+1})}{1- r}\). Here, a= 2 and r= 1/3 with n= k- 1 so that is \(\displaystyle \frac{2(1- (\frac{1}{3})^{k})}{1-\frac{1}{3}}= (3/2)(1- (\frac{1}{3})^{k})\). Subtract that from 1 to get \(\displaystyle P(X\ge k)\).