Is answer 2/45?

given 8/125. My own 8/153.

My solution (need to be verified so I didn’t post as ans to attract views)
(1/3)^3 [1+ (1/6)(1/2) + (1/6)(1/2)^2 + (1/6)^2(1/2)^2 + (1/6)^2(1/2)^3 ...] =
(1/3)^3 [1 + 7/24 + 7/156 + 7/13824...] =
(1/3)^3 [1 + (7/24)/(1- 1/24)] =
8/153

Rationale: Counting from the end of the infinite sequence. 1 is the shortest sequence possible of rolling 3 or 5 consecutively. 1/6 accounts for rolling of 1. (1/2)^2 accounts for longest possible roll on 3 or 5 (roll on 1 is also summed within). Subsequent 1/6 is to break chain. the powers follow a geometric sequence when grouped into 3.

https://math.stackexchange.com/questions/1646860/probability-of-a-fair-sequence-of-tosses-ending-on-two-successive-tails-given-th

very similar but less complicated question solved using Markov chain. (Google search link or use website then copy paste)

Is there any way in part a or b show formula to do summation of (r+1)(r+1)/(6^r)? I used GC I got 0.06399999, which is 8/125...

Nope, you could refer to the full question posted as “answer”