Titration of barium(II) hydroxide by hydrochloric acid
Your solution is correct up to the point you assumed that you can double the concentration of hydrochloric acid.
Unfortunately, this is wrong and not at all how stoichiometry works.
Let's focus on what's important.
In the nutshell, we are dealing with a typical neutralization reaction:

$$\ce{H3O+ + OH- &lt;=&gt; 2 H2O}$$

and note that $\ce{BaCl2}$ as a salt of a strong acid and a strong base won't undergo hydrolysis and, as a consequence, won't have a noticeable impact on $\mathrm{pH}$.

As you [semi-]correctly calculated, there is an excess of a base ($n_0$ values refer to the initial amounts):

$$
\begin{align}
n_0(\ce{H3O+}) &amp;= n(\ce{HCl}) \\
  &amp;= c(\ce{HCl})\cdot V(\ce{HCl}) \\
  &amp;= \pu{2.4e-2 mol L-1}\cdot\pu{2.5e-2 L} \\
  &amp;= \pu{6e-4 mol}
\end{align}
$$

$$
\begin{align}
n_0(\ce{OH-}) &amp;= 2\cdot n(\ce{Ba(OH)2}) \\
  &amp;= 2\cdot c(\ce{Ba(OH)2})\cdot V(\ce{Ba(OH)2}) \\
  &amp;= 2\cdot\pu{3.4e-2 mol L-1}\cdot\pu{4.5e-2 L} \\
  &amp;= \pu{3.06e-3 mol}
\end{align}
$$

One can see that there is a tremendous excess of hydroxide of the following amount:

$$
\begin{align}
n(\ce{OH-}) &amp;= n_0(\ce{OH-}) - n_0(\ce{H3O+}) \\
  &amp;= \pu{3.06e-3 mol} - \pu{6e-4 mol} \\
  &amp;= \pu{2.46e-3 mol}
\end{align}
$$

and the leftover concentration is

$$
\begin{align}
c(\ce{OH-}) &amp;= \frac{n(\ce{OH-})}{V(\ce{HCl}) + V(\ce{Ba(OH)2})} \\
  &amp;= \frac{\pu{2.46e-3 mol}}{\pu{2.5e-2 L} + \pu{4.5e-2 L}} \\
  &amp;= \pu{3.51e-2 mol L-1}
\end{align}
$$

Finally, as you rightfully suggested (use the correct value for $c(\ce{OH-})$, however):

$$
\begin{align}
\mathrm{pH} &amp;= 14 - \mathrm{pOH} \\
  &amp;= 14 + \log(\pu{3.51e-2}) \\
  &amp;= 12.55
\end{align}
$$

Notes


Using the term "moles of X" is incorrect.
The proper terminology here is the amount of substance.
Never try to substitute a physical quantity with its unit.
Watch out for the proper capitalization, which is actually a big deal in natural sciences.
$\ce{HCl}$ is a formula for hydrogen chloride, whereas $\ce{HCL}$ is not and may denote "hydrogen-carbon-ligand" or be an abbreviation.
Same goes for math operators, including $\mathrm{pH}$ and $\mathrm{pOH}$: $\ce{Ph}$ is phenyl and $\ce{PH}$ can be read as "phosphor-hydrogen", or be yet another abbreviation.