I really need help with that question, so I'd appreciate a solution with steps because this problem was given for practice and my exam is in a few days so I'd like to understand the steps to the solution. Thank you.
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I think what this question is getting at is that, to be a basis of a vector space,

a set of vectors must span the vector space and be linearly independent.

If $\pmatrix{x\\y}$ is any element of $\mathbb R^2$, we can solve $\pmatrix{x\\y}=a\pmatrix{1\\2}+b\pmatrix{1\\3}$

to get $a=3x-y$ and $b=y-2x$,

so $\pmatrix{x\\y}$ is a linear combination of $\pmatrix{1\\2}$ and $\pmatrix{1\\3}$, so $\pmatrix{1\\2}$ and $\pmatrix{1\\3}$ span $\mathbb R^2$

(so certainly $\pmatrix{1\\2}$, $\pmatrix{1\\3},$ and $\pmatrix{1\\4}$ span $\mathbb R^2$).

Indeed, any two linearly independent vectors in $\mathbb R^2$ (which is a $2$-dimensional vector space over $\mathbb R$)

will span $\mathbb R^2$ and be a basis.

Elements of a basis must be linearly independent, but $(1,4)=2(1,3)-(1,2)$,

so the set of three vectors given are not a basis.
**HINT**

To spell out the comment hint a bit more, if $U$ spans $\mathbb{R}^2$, and $\vec{b} \in \mathbb{R}^2$, then there must exist $x,y,z \in \mathbb{R}$ such that
$$
\vec{b}
 = x\begin{bmatrix} 1\\2\end{bmatrix}
 + y\begin{bmatrix} 1\\3\end{bmatrix}
 + z\begin{bmatrix} 1\\4\end{bmatrix}
 = \begin{bmatrix} 1 & 1 & 1 \\2 & 3 & 4\end{bmatrix}
 \begin{bmatrix} x\\y\\z\end{bmatrix}
$$
So you have to prove that any $b$ can be generated that way...