Is photosynthesis endothermic and exothermic at the same time? ... somehow?
Consider that when $1$ molecule of glucose forms, this happens at the cost of $6$ molecules of  $\ce{CO2}$ and $6$ molecules of $\ce{H2O}$ :

$
\ce{6CO2(g) + 6H2O(l) -&gt; C6H12O6(s) + 6O2(g)}
$

Considering that at $25\ \mathrm{^\circ C}$ and  $1\ \text{atm}$

$
\Delta H_{\text{f},\ce{CO2}}^\circ=-393.5\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{H2O}}^\circ=-285.8\ \text{kJ/mol}
$

$
\Delta H_{\text{f},\ce{C6H12O6}}^\circ=-1273.3\ \text{kJ/mol}
$

You get, by summing the enthalpies of formation for each of the involved species and considering that for $\ce{O2}$ that enthalpy is zero:

$
\Delta H_\text{r}^\circ=6\times393.5\ \text{kJ/mol}+6\times285.8\ \text{kJ/mol}-1273.3\ \text{kJ/mol}=2802.5\ \text{kJ/mol}
$

Note that, for $\ce{CO2}$ and $\ce{H2O}$, the change of sign is required, since you "lose" them, while you form glucose.

So, finally $\Delta H_\text{r}^\circ$ (heat to be absorbed to form one mole of glucose) is positive, indicating that photosynthesis is indeed endothermic.