For the dilution, just remember that M (molarity) has units of moles per liter (mol/L), so you can find concentration (M), volume (L), or moles in solution (mol) with this relationship: M = mol/L (you might see this more generally as concentration = moles / volume, or c = m/v)

We have M = 12 M and L = 3.0 L, so: 12 M = mol/3.0 L --> rearranged to solve for mol: mol = 12 M * 3.0 L = 36 moles in solution.

Now we can use our target concentration, 3.25 M, to find what volume we must dilute to:

3.25 M = 36 mol / L --> solving for L: L = 36 mol/3.25 M = ~11.1 L

For the pH problems, you need to remember these definitions:

pH = -log[H^{+}] (and thus [H^{+}] = 10^{-pH})

pOH = -log[OH^{-}] (and thus [OH^{-}] = 10^{-pOH})

pH + pOH = 14

So we can find the [H^{+}] right away by solving [H^{+}] = 10^{-9.2} = 6.3x10^{-10} M

We need pOH to find [OH^{-}], so we can subtract 9.2 from 14 (because of this definition: pH + pOH = 14)

14 - 9.2 = pOH = 4.8

then [OH^{-}] = 10^{-4.8} = 1.59x10^{-5}

Now this last question should be straightforward: we are taking the negative log of the given concentration

pH = -log[H^{+}]

The calculator button will say LOG, and my calculator has a (-) button to make it negative. You also may have an EE button to make scientific notation easier. This is what my entry looks like:

-log(2.7E-7) = 6.568636236

and that will be 6.6 when rounded to 2 sigfigs.

Hope this helps!