Your approach is all right but the solution given by the textbook is wrong :), at least if no approximation is to be made.
Let's go the other way around: start from $$P(V,T) = \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2V_0}\left[\left(\frac{V_0}{V}\right)^5 - \left(\frac{V_0}{V}\right)^3\right]$$ and then derive the values of $\kappa_T$ and $\beta$ (by the way, shouldn't it rather be $\chi_T$ (see here) and $\alpha$ (here)?). As you mentioned, to do this, we need to compute partial derivatives of $P$:
$$
\left(\frac{\partial P}{\partial T}\right)_V = \frac{\Gamma c_v}{V}
$$
$$
\left(\frac{\partial P}{\partial V}\right)_T =
-\left(\frac{\Gamma c_vT}{V^2} + \frac{\varepsilon}{2{V_0}^2}\left[5\left(\frac{V_0}{V}\right)^6 - 3\left(\frac{V_0}{V}\right)^4\right]\right)
$$
which give
$$
\frac{1}{\kappa_T} = -V \left(\frac{\partial P}{\partial V}\right)_T
= \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2{V_0}}\left[5\left(\frac{V_0}{V}\right)^5 - 3\left(\frac{V_0}{V}\right)^3\right]
$$
and
$$
\beta = \kappa_T \left(\frac{\partial P}{\partial T}\right)_V
= \frac1T \left(1 + \frac{\varepsilon }{2\Gamma c_V T}\left[5\left(\frac{V_0}{V}\right)^4 - 3\left(\frac{V_0}{V}\right)^2\right]
\right)^{-1}
$$
Clearly, that's not what the textbook gives in the first place, but it's close enough to understand what they did: a Taylor expansion at order 3 in $V_0/V$ in booth cases, which gives back the expressions
$$
\frac{1}{\kappa_T}
\approx \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2{V_0}}\left[ - 3\left(\frac{V_0}{V}\right)^3\right]
$$
and
$$
\beta
\approx \frac1T \left(1 + \frac{\varepsilon }{2\Gamma c_V T}\left[ - 3\left(\frac{V_0}{V}\right)^2\right]
\right)^{-1}
\approx \frac1T \left(1 + \frac{\varepsilon }{2\Gamma c_V T}\left[ 3\left(\frac{V_0}{V}\right)^2\right]
\right)
$$
It really should have been clearer in the text that you could suppose $V\gg V_0$ (and not only $V>V_0$) and I don't see how you could derive the term in $\left(\frac{V_0}{V}\right)^5$ from this as it is completely neglected to find back $\kappa_T$ and $\beta$
For a given process, the heat added divided by the temperature change of the system (I am assuming they are calling this C) varies with the amount of work that is done. Like you, I can't see why they would possibly call this an intensive property, although it is certainly a path function. Maybe intensive is a typo, and they meant extensive.
The specific heat capacities Cv and Cp are intensive state functions, because they are defined as the partial derivatives of the specific internal energy and the specific enthalpy, respectively, with respect to temperature (the former at constant volume and the latter at constant pressure), and the specific internal energy and specific enthalpy are state functions.
Best Answer
The key is to note the left-hand terms: $v(T,p)$ and $u(T,v)$. The presence of two variables reminds us that for a closed system, we can add energy in two ways: work or heat. The implication is that any state variable can be expressed in differential form using at most two other state variables.
We start accordingly by expanding $dv$ in the variables $T$ and $p$ as
$$dv(T,p)=\left(\frac{\partial v}{\partial T}\right)_pdT+\left(\frac{\partial v}{\partial p}\right)_Tdp;$$
$$dv(T,p)=v\beta(T)\,dT-v\kappa(T)\,dp;$$
$$\frac{dv(T,p)}{v}=\beta(T)\,dT-\kappa(T)\,dp,$$
where we've used the definitions of the material properties $\beta$ (the thermal expansion coefficient) and $\kappa$ (the compressibility). Assuming temperature-independent material properties; integrating from $v_0$ to $v$, $T_0=0$ to $T$, and $p_0=0$ to $p$; and assuming small changes in volume so that we can Taylor-expand $\ln \frac{v}{v_0}\approx\frac{v-v_0}{v_0}$, we obtain the first equation you provide.
For the second equation, we expand $du$ ($=T\,dS-p\,dv$ for a closed system) in $T$ and $v$ as
$$du(T,v)=\left(\frac{\partial u}{\partial T}\right)_vdT+\left(\frac{\partial u}{\partial v}\right)_Tdv;$$
$$du(T,v)=c_v(T)\,dT+\left[T\left(\frac{\partial S}{\partial v}\right)_T-p\right]dv;$$
$$du(T,v)=c_v(T)\,dT+\left(\frac{T\beta(T)}{\kappa(T)}-p\right)dv,$$
where we've used a Maxwell relation and the material property definitions again. Now assume temperature-independent $\beta$ and $\kappa$ again, use $\frac{\beta T}{\kappa}=\frac{1}{\kappa}\left(\frac{v-v_0}{v_0}\right)+p$ from the first equation, integrate from $v_0$ to $v$, and assume $u_0=0$ to obtain the second equation you provide.