75 ml of a solution of HNO3 (0.0125 M) are added to 130 ml of a solution of Be(OH)2 (0.0240 M). Calculate the ph of the final solution and the volume of HCl or Mg(OH)2 to add to neutralize the obtained (final) solution. The concentration of the acid or base for the neutralization is 0.075.

This is a classic acid/base chemistry question.  First let's find the total number of moles of protons and hydroxide ions that will be present in the reaction.  First, we multiply the molarity of HNO3 (nitric acid) by the volume:
(0.0125 moles / L) * 0.075 L = 0.0009375 moles of H+
Now let's do the same thing to find the moles of hydroxide (OH-) present.  Each mole of Be(OH)2 will produce 2 moles of OH- so we need to multiply by 2:
(0.0240 moles / L) * 0.130 L * 2 = 0.00624 moles of OH-
Since the H+ will react with OH- to produce H2O, we need to see which species in present in a greater amount.  There are more moles of hydroxide present than protons, so we can subtract the numbers to find out what the excess amount of hydroxide remaining is:
0.00624 - 0.0009375 = 0.0053025 moles of OH- remaining.
We can divide the moles of hydroxide by the total volume of solution present to find the concentration of hydroxide.  The total volume of the solution is 130 mL + 75 mL = 205 mL
0.0053025 moles / 205 mL = 0.0259 M
We can convert the molarity of hydroxide ions into a pOH value by taking the negative log of the concentration via the equation below:
pOH = -log[OH-] = -log(0.0259) = 1.59
Since pH + pOH = 14, that means that we subtract the pOH from 14 to find the pH:
pH = 14 - pOH = 14 - 1.59 = 12.41
So the pH of the solution is 12.41.  The amount of a 0.075 M solution of HCl needed to add to neutralize the hydroxide is calculated below:
0.0053025 moles * (1 L / 0.075 moles) = 0.0707 L = 70.7 mL
So 70.7 mL of HCl solution would be required to neutralize the resulting solution.