Hint: You correctly translated "odd function". You know $f(3)$, so you know $f(-3)$ and similarly for any couple of opposite values of the argument of $f$.
Now, $f$ is periodic of period $10$, so $f(3)=f(3-10)=f(-7)$. But then...
Due to symmetry we need to define $f(x)=x^2+2$ on $(-1,0)$. It remains to discuss the $2$-periodicity. One starts with "special points", for example $x=0$. Then
$$f(0)=0^2+2=\text{(2-periodicity)}=f(0+2)=f(2),$$
i.e. $f(2):=2$. To find the extension of $f$ on $(2,3)$ one continues with all $\delta\in (0,1)$, i.e.
$$f(\delta)=\delta^2+2=\text{(2-periodicity)}=f(\delta+2),$$
where $\delta+2\in (2,3)$. Similarly, considering all points $-\delta\in(-1,0)$, using $2$-periodicity one finds the extension of $f$ on the interval $(1,2)$ via $-\delta+2\in (1,2)$, as we did above. The extension of $f$ for all other points easily follow by drawing.
Due to symmetry we need to define $f(x)=-x^2-2$ on $(-1,0)$. One selects once again
the "special points", for example $x=0$ as in the even case. The analysis is completely similar, with due changes.
Best Answer
the function is periodic with period $T$ and $f(x)+f(-x)=0$, you can combine the fact that $f(x+T)=f(x)$ and then
$$\int_{-T/2}^{T/2}f(x)dx=0\\ \int_{-\frac{T}{2}}^{\frac{T}{2}}f(x)dx=\\ \int_{-\frac{T}{2}}^{0}f(x)dx+\int_{0}^{\frac{T}{2}}f(x)dx\\ \int_{-\frac{T}{2}}^{0}f(x+T)dx+\int_{0}^{\frac{T}{2}}f(x)dx\\ y=x+T\\ dy=dx\\ \int_{\frac{T}{2}}^{T}f(y)dy+\int_{0}^{\frac{T}{2}}f(x)dx=\\ \int_0^Tf(x)dx$$
now you can conclude you argument.