How to optimally solve and generalize this Conway's Game of Life puzzle

My "Game of Life" implementation:

a = RandomInteger[BernoulliDistribution[0.1], {200, 200}];

Dynamic[Image[a =
   1 - Unitize[(# - 3) (# - 12) (# - 13)] &@
    ListConvolve[{{1, 1, 1}, {1, 10, 1}, {1, 1, 1}}, a, {2, 2}, 0]
  ]]

To use the initial grid in the question you would just initialise a with:

a = SparseArray[list -> 1]

However my code continuously updates the grid (quite rapidly), so the grid evolves to the "all dead" state in a fraction of a second.

To actually answer the question "how many generations have any life in them" you could write the evolution step as a function and use FixedPointList to count the number of steps:

evolve[a_] := 1 - Unitize[(# - 3) (# - 12) (# - 13)] &@
  ListConvolve[{{1, 1, 1}, {1, 10, 1}, {1, 1, 1}}, a, {2, 2}, 0]

Length@FixedPointList[evolve, Normal[SparseArray[list -> 1]]] - 2
(* 3 *)

Note that FixedPointList will generate the complete list of grids, so for larger problems you would want to use a different approach to avoid eating all your computer's memory.

This is one way of implementing Conway's Game of Life:

step[matrix_] := Module[{m = matrix, neighbours},
  neighbours = ListConvolve[{{1, 1, 1}, {1, 0, 1}, {1, 1, 1}}, ArrayPad[m, 1]];
  m = ReplacePart[m, Position[neighbours, _?(# == 1 || # > 3 &)] -> 0];
  m = ReplacePart[m, Position[neighbours, 3] -> 1]
  ]

For example, this is the so called glider:

m = ArrayPad[{{0, 1, 0}, {0, 0, 1}, {1, 1, 1}}, {1, 10}];
ListAnimate[ArrayPlot[#, Mesh -> True] & /@ NestList[step, m, 30]]

I've written about another cellular automata problem here and this includes a link to a post where image processing was used in Mathematica to get initial states for a cellular automata problem, so the answer to that question is that yes it is possible.