If (1+m²)x² - 2(1+3m)x + (1+8m) = 0. Then numbers of value(s) of m for which this equation has no solution.

We know that D = b² - 4ac
D = 4(1+3m)² - 4(1+m²)(1+8m) = 4(1 + 9m² + 6m - (1 + 8m + m² + 8m³)) = 4(8m² - 2m - 8m³) = -8(4m³ - 4m² + m) = -8m(4m² - 4m + 1) = -8m(2m - 1)² < 0.
Since (2m - 1)² is always positive, for D < 0, we must have -8m < 0 which implies m > 0.
Hence infinitely many values.