Assuming that 3 are true, the only self-consistent set are 1,2 and 3.
Importantly, statements 3 and 4 contradict one another, so one of them must be the incorrect statement.
Consider a reaction with a change in enthalpy ($\Delta H$) of $-500$ kJ/mol$\pu{-500 kJ/mol}$, and statement 3 is true:
$$\Delta H = H_f - H_i\\ \implies H_i = H_f-\Delta H\\ H_i= H_f + 500 \text{ kJ/mol}\\ \implies H_i>H_f $$\begin{align} \Delta H &= H_\mathrm{f} - H_\mathrm{i}\\ \implies H_\mathrm{i} &= H_\mathrm{f} - \Delta H\\ H_\mathrm{i} &= H_\mathrm{f} + \pu{500 kJ/mol}\\ \implies H_\mathrm{i} &> H_\mathrm{f} \end{align} So we'veWe've shown that statement 1 is true also.
As wolphram says in the comments, you can't measure an absolute value of enthalpy, but you can measure it relative to an chosen zero enthalpy. If you chose zero enthalpy such that $H_i=1600$ kJ/mol$H_\mathrm{i} = \pu{1600 kJ/mol}$, then by the first equation above we can say that: $$ H_f = \Delta H + H_i\\ \implies H_f = -500 + 1600\\ \implies H_f= 1100 \text{ kJ/mol} $$\begin{align} H_\mathrm{f} &= \Delta H + H_\mathrm{i}\\ \implies H_\mathrm{f} &= \pu{-500 kJ/mol} + \pu{1600 kJ/mol}\\ \implies H_\mathrm{f} &= \pu{1100 kJ/mol} \end{align} So we'veWe've shown that statement 2 is true as well.
Statement 4 contradicts statement 3 (as well as the other two), so statement 4 must be incorrect.