Calculate the signal's average power We have this signal:

$$x(t)= 2 + 2\cos(2\pi f_0 t) + \frac{2}{T_0} \operatorname{sinc}\left(\frac{2 t}{T_0}\right)e^{j2 \pi 4/ T_0} + \operatorname{sinc}\left(\frac{2 t}{T_0}\right)e^{-j2 \pi 4/ T_0}.$$

I have calculate Fourier trasformed and have found:
$$
X(f)=\delta(f)+ \delta(f-f_0) + \delta(f+f_0) + \operatorname{rect}\left(\frac{f-\frac 4{T_0}}{\frac 2{T_0}}\right) + \operatorname{rect}\left(\frac{f+\frac 4{T_0}}{\frac 2{T_0}})\right)
$$

I must find the average power. If use Parseval's have:
\begin{align}
P_x&amp;=\frac{1}{T_0} \int_{-\infty}^{\infty}\lvert X(f)\rvert^2df \\
&amp;= \frac 1{T_0}\left(\int\lvert\delta(f)+\delta(f+f_0)+\delta(f−f_0)+\operatorname{rect}+\operatorname{rect}\rvert^2df\right).
\end{align}
How can I solve this?
A signal either has finite energy, finite power or even infinite power. If it has finite energy, it will have zero average power, according to your definition
$$P_x=\lim_{T_0\rightarrow\infty}\frac{1}{T_0}\int_{-\frac{T_0}{2}}^{\frac{T_0}{2}}|x(t)|^2dt.$$

Knowing that the sincs are orthogonal to each other, as well as the cos function, the average power is given by

$$P_x=P(2)+P(2\cos(2\pi f_0t))+P(\operatorname{sinc})$$

where $P(...)$ is the power of the components in brackets. The sinc function has finite energy, hence it has zero power on average. Hence, we have 

$$P_x=P(2)+P(2\cos(2\pi f_0 t)$$
with
$$P(2)=4$$
and $$P(2\cos(2\pi f_0 t)=2^2\cdot\frac{1}{2}=2.$$

Therefore, the overall power is given by

$$P_x=6.$$

For the calculation of the $\cos$, consider
$$P_{\cos}=\frac{1}{T}\int_{0}^{T}\cos^2(2\pi t/T)dt=\frac{1}{2}$$