Because c and m only occur as c^m, you can only deal with c^m and can't get separate estimates for c and m. Also if you change 0.75 to 3/4 and set t to specific values, a solution for c^m, a, and b occurs.
t = 4;
eq1 = 1 == b*cm + a
eq2 = 3/4 == b*cm^(t) + a
eq3 = 0 == b*c^(0) + a
Reduce[{eq1, eq2, eq3}, {a, b, cm}] // ToRadicals
(a == 1/13 (8 + (2 (81 + 13 Sqrt[57]))^(1/3)/3^(2/3) - (8 2^(2/3))/(3 (81 + 13 Sqrt[57]))^(1/3)) ||
a == 8/13 - ((1 + I Sqrt[3]) (81 + 13 Sqrt[57])^(1/3))/(13 6^(2/3)) + (4 2^(2/3) (1 - I Sqrt[3]))/(
13 (3 (81 + 13 Sqrt[57]))^(1/3)) ||
a == 8/13 - ((1 - I Sqrt[3]) (81 + 13 Sqrt[57])^(1/3))/(13 6^(2/3)) + (4 2^(2/3) (1 + I Sqrt[3]))/(
13 (3 (81 + 13 Sqrt[57]))^(1/3))) && b == -a &&
cm == 1/4 (-12 + 24 a - 13 a^2)
