Infinite Increase in Entropy when Energy added to Absolute Zero
The textbook is referring to the entropy change of the system.  While the textbook is correct that absolute zero can never be attained, its statement that the entropy change is infinite is wrong. The authors' rationale for thinking it is infinite likely stems from a misinterpretation of the definition of entropy change:
$dS = \frac{\text{đ}q_{rev}}{T}$
where $\text{đ}q_{rev}$ is the reversible heat flow into the system.
Let's start with a system at $0 K$, and then warm it to a temperature $T = T'$.  Then:
$\Delta S =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$
Let's consider what happens when we flow the first, infinitesimal amount of heat, $\text{đ}q$, into the system.  And to make the calculation easy, let's assume the heat flow is reversible (it doesn't have to be; if it weren't, we'd just need to find a reversible process that gets us to the same final state).  As soon as we do this, the temperature is no longer zero! And once the temperature rises above zero, the singularity disappears and we no longer have a concern about infinite entropy change.
But, you may ask, what about the mathematics right at the very beginning, when the temperature is indeed zero.  Here, to properly calculate the entropy change, we need to use a limit, recognizing that both $T$ and $\text{đ}q$ are approaching zero.  The easiest way to understand this is to consider the Debye $T^3$ law, which models the constant-volume heat capacity of solids as they approach absolute zero:
$C_v = C_v(T) = k T^3$, where $k$ is a constant, and where I've written &quot;$C_v(T)$&quot; as an explicit indicator that $C_v$ is temperature-dependent.
Since, for a constant-volume reversible process, $\text{đ}q_{rev} = C_v(T) dT$, we have:
$\Delta S= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T} = \int_{0}^{T'}\frac{C_v(T)}{T} dT = \int_{0}^{T'}\frac{k T^3}{T} dT=\int_{0}^{T'}k T^2 dT$
I.e., in evaluating what happens to $dS = \frac{\text{đ}q_{rev}}{T}$ as $T \rightarrow  0$, it is necessary to consider what is happening to both the numerator and the denominator. The textbook authors' mistake was in not understanding this.  Clearly, once we find a reasonable functional form for $\text{đ}q$ in the limit as $T \rightarrow  0$, the singularity disappears.
Think of it this way (borrowing from one of my comments): At constant volume, $dS=\frac{C_v}{T} dT$. As you approach absolute zero, it's not just $T$ that's going to $0$; $C_v$ is going to $0$ as well. And since $C_v$ is going to $0$ faster than $T$ is going to $0$, $\frac{C_v}{T}$ goes to $0$ rather than infinity (for the same reason that $\frac{x^3}{x}$ goes to $0$ rather than infinity as $x$ goes to $0$). I.e., the mistake would be in assuming that $C_v$  is a constant; it's not.
Here's another way we can understand that the textbook's statement is wrong:  Thermodynamics allows us, in principle, to determine absolute entropies for any substance.  The absolute entropy is given by integrating the entropy change from absolute zero to whatever temperature the substance is at:
$\text {Absolute entropy at } T' \equiv S(T') =\Delta S_{0 \rightarrow T'}= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$
If the entropy change between $0 K$ and any higher temperature were infinite, the absolute entropies of all real substances would likewise also be infinite! [Unless the substance were at absolute zero, which is unatainable.]
Here's a link showing how standard molar entropies might be calculated (I don't know if the procedure described here is what is used to determine the official CODATA values):
https://www2.stetson.edu/~wgrubbs/datadriven/entropyaluminumoxide/entropyal2o3wtg.html.
Essentially, it mentions using the Debye extrapolation up to ~$15 K$, and then using a differential scanning calorimeter to determine the reversible heat flow between $15 K$ and $298.15 K$