HClO4(s) -> HClO4(aq)
n (HClO4) = m / mm
n (HClO4) = 2.512 / 100.458
n (HClO4) = 0.025 mol
C (HClO4) = 0.025 / 1
C (HClO4) = 0.025 molL^-1
pH = - log [H+]
pH = - log [0.025]
pH = 1.6
Now for (b) the concentration decreases by a factor of 10, so the pH rises 1.
pH = 2.6