Find all solutions of the linear least squares problem $$\min_x\|Ax-b\|_2$$ where $A = \left[\begin{matrix} 1&1\\1&1\end{matrix}\right]$ and $b = \left[\begin{matrix} 1 \\ 2\end{matrix}\right]$.
Note that using the SVD of $A = U\Sigma V^T$ we find $A^+$ (the Moore-Penrose pseudoinverse) finding $A^+ = V\Sigma^+U^T$, where for $\alpha \in\mathbb R $, $\alpha^+ =\begin{cases} \alpha^{-1}, &\alpha \ne 0 \\ 0, &\alpha = 0\end{cases}$. Then $A^+ = \left[ \begin{matrix} \frac 14 & \frac 14 \\ \frac 14 & \frac 14 \end{matrix} \right]$.
Then I think it is asking we can find $x = A^+ b = \left[\begin{matrix} \frac 14 & \frac 14 \\ \frac 14 & \frac 14 \end{matrix}\right]\left[\begin{matrix} 1 \\ 2\end{matrix}\right] = \left[\begin{matrix} \frac 34 \\ \frac 34\end{matrix}\right] $.
Is there more to the question that I am missing?
You are missing that your solution is not unique.
You have found a valid least-squares solution, but note that $A$ has a nontrivial kernel: it takes, for example, $v = (1,-1)$ to $0$. Accordingly, we can add any element of the kernel to our solution to obtain another valid solution.
It is fairly evident that the kernel of $A$ consists of scalar multiples of the particular $v$ we tried above, so our full solution is
$x + av$
where $x$ is as you have computed, $v$ is as above, and $a$ is some scalar.