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Target equation: $\ce{Mg(s) + 1/2 O2 -> MgO(s)}$
$$ \begin{align} \ce{MgO(s) + 2 HCl(aq) &-> MgCl2(aq) + H2O(l)} &\quad ΔH &= \pu{-1300 kJ} \tag{1}\\ \ce{Mg(s) + 2 HCl(aq) &-> MgCl2(aq) + H2(g)} &\quad ΔH &= \pu{-602 kJ} \tag{2}\\ \ce{H2(g) + 1/2 O2 &-> H2O(l)} &\quad ΔH &= \pu{-286 kJ} \tag{3} \end{align} $$
The target equation should have a negative $ΔH$ meaning that it's exothermic, but when I add it up, $ΔH$ ends up being positive. Is there anything I'm missing?
Your target equation is: $\ce{Mg(s) + 1/2O2 -> MgO(s)}$. You have given following set of equations to get your target's enthalpy of formation: $$\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)} \quad ΔH = \pu{-1300 kJ} \tag {1}$$ $$\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \quad ΔH = \pu{-602 kJ} \tag {2}$$ $$\ce{H2(g) + 1/2O2 -> H2O(l)} \quad ΔH = \pu{-286 kJ} \tag {3}$$
Yet, if you added equations $(1)$ through $(3)$, you won't get your target equation because you have set equation $(1)$ in opposite direction so that $\ce{MgO}$ would ends up in reactant side (LHS). What you should have done is reverse that reaction. The reverse of Equation $(1)$ has identical $\Delta H$ but with opposite sign:
$$\ce{MgCl2(aq) + H2O(l) -> MgO(s) + 2HCl(aq)} \quad ΔH = \pu{+1300 kJ} \tag {4}$$
Now you can add equations $(2)$ through $(4)$, and you should get a negative $ΔH$ for your target equation (meaning the reaction is exothermic, and given $\Delta H^\circ_f$ for $\ce{MgO}$ is $\pu{−601.6 kJ/mol}$ according to Wikipedia). However, your given $\Delta H$ values are seemingly incorrect and I leave you to find the correct values (Note: The given $\Delta H^\circ_f$ for water is acceptable).
Late edition: According to OP's late comment to andselisk's suggestion on first comment, it seems OP has made a mistake on amount of $\ce{MgO}$ calculation, of which OP has used in the calarimetric experiment. It should have been $\pu{\approx 0.025 mol}$, if OP has used $\pu{\approx 1 g}$ of $\ce{MgO}$. Thus, $\Delta H$ for the reaction $\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)}$ should have been $\approx \pu{3.34 kJ}/\pu{0.025 mol}= \approx \pu{134 kJ/mol}$. Thus, $\Delta H^\circ_f$ of $\ce{MgO}$ from the experiment is: $\pu{(+134-602-286) kJ/mol}=\pu{-754 kJ/mol}$ (literature value is $\pu{−601.6 kJ/mol}$).
Until the time of this edition, OP has not edited the question. Thus, I assume OP has also calculated the data for the reaction $\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}$, actual value of which has been around $\pu{−460 kJ/mol}$. Hence, approximate experimental value for $\Delta H^\circ_f{_\ce{MgO}}$ is: $\pu{(+134-460-286) kJ/mol}=\pu{-612 kJ/mol}$, which is better acceptable value for literature value, $\pu{−601.6 kJ/mol}$.
Why do you add the three equations ?
By first adding (2) and (3) and then SUBSTRACTING (1), you find Mg + (1/2) O2 --> MgO.
On the other end, I think some of these numerical values are wrong, because one should obtain -602 kJ/mol, which is the heat of formation of MgO. And the combination of the given values do not yield to this result, by far.
