A spacecraft moves in a circular orbit of radius r0 around a planet, with period T0. A short thruster burn along its direction of motion increases…

Let $v_0^2=GM/r_0$. After the burn the speed at perigee is $v_p^2=\tfrac74v_0^2$. Energy conservation with angular momentum conservation ($v_pr_0=v_ar_a$) gives the vis-viva relation at perigee: $v_p^2=\dfrac{2GM\,r_a}{r_0(r_0+r_a)}$, i.e. $\dfrac74=\dfrac{2r_a}{r_0+r_a}$, so $\tfrac74r_0=\tfrac14r_a$ and $r_a=7r_0$. The semi-major axis is $a=\dfrac{r_0+7r_0}{2}=4r_0$. By Kepler's third law $T^2\propto a^3$: $T=T_0\left(\dfrac{4r_0}{r_0}\right)^{3/2}=8T_0$. Perigee to apogee is exactly half of a period: $t=4T_0$. Using $a=r_a=7r_0$ or $a=r_0$ is a typical slip. Kepler's second law is not needed: the time from perigee to apogee is half the period for any eccentricity, so only the semi-major axis matters.