The numbers are telling you what representations should you use to transform the fields. Let's see an example:
As we know the Dirac Lagrangian can be written as $$\mathscr{L}=\bar{\psi}\left(i\partial_\mu\gamma^\mu-m\right)\psi$$. Is easy to see that under a local transformation of group ##U(1)##, ##\psi'=e^{igY\theta}\psi## the Lagrangian is invariant. But note that ##Y## is an arbitrary number (a diagonal matrix if you work with more than one field). So for each field, you can assign a different value of ##Y##. That's the Hypercharge and it's the 3rd number of your table, and it's an input of SM, you cannot deduce it.
For the first two numbers is the same idea, imagine that, instead of one field you have two (or three) fields with the same mass, then you can write the Dirac Lagrangian as
$$\mathscr{L}=\begin{pmatrix}\bar{\psi}_1&\bar{\psi}_2\end{pmatrix}
\left(i\partial_{\mu}\begin{pmatrix}\gamma^\mu&0\\0&\gamma^\mu\end{pmatrix}-\begin{pmatrix}m&0\\0&m\end{pmatrix}\right)\begin{pmatrix}\psi_1\\\psi_2\end{pmatrix}$$
Now the Lagrangian is invariant under transformations of the group ##SU(2)##; $$\psi'=S\psi \qquad S=e^{ig_W\vec{T}\cdot \vec{\theta}}$$. But ##S## can belong to any representation of ##SU(2)##, so again we have an arbitrary choice of ##\vec{T}##. And that's what the second column of you table tell you. For those that transform with the trivial representation (1) ##\vec{T}=0\Longrightarrow S=I##, but for those that transform with the fundamental representation (2) ##\vec{T}=\frac{\vec{\sigma}}{2}## where ##\sigma## are the Pauli matrices. The same can be done with ##SU(3)##.