$$P(x)=x^4+ax^3+bx^2+cx+d$$ The sum of pair of roots is given by $1,2,5,6,9,10$
I solved it by letting the roots $=\alpha,\beta,\gamma,\delta$ and then find them individually.
But since we have six equations and 4 variables, there's always a chance that system might become inconsistent.
Is there some way to do it without finding $\alpha,\beta,\gamma,\delta$ and just using $a,b,c,d$ ?
PS
$\alpha=-3/2$
$\beta=5/2$
$\delta=7/2$
$\gamma=13/2$
P$(1/2)=-72$
As you wrote, it is possible that the system is inconsistent.
To check if it is (or not), let us define
$$ f_1=\alpha+\beta - 1 $$ $$ f_2=\alpha+\gamma- 2$$ $$ f_3=\alpha+\delta - 5 $$ $$ f_4=\beta+\gamma-6 $$ $$ f_5=\beta+\delta - 9 $$ $$ f_6=\gamma+\delta -10$$
and define a norm $$F=\sum_{i=1}^6 f_i^2$$
Compute the partial derivatives (this now leads to a square system) $$\frac{\partial F}{\partial \alpha}=6 \alpha +2 \beta +2 \gamma +2 \delta -16$$ $$\frac{\partial F}{\partial \beta}=2 \alpha +6 \beta +2 \gamma +2 \delta -32$$ $$\frac{\partial F}{\partial \gamma}=2 \alpha +2 \beta +6 \gamma +2 \delta -36$$ $$\frac{\partial F}{\partial \delta}=2 \alpha +2 \beta +2 \gamma +6 \delta -48$$
Set each of them equal to $0$ since we look for the minimum of the norm $F$. This gives $$\alpha= -\frac{3}{2}\qquad \beta = \frac{5}{2}\qquad \gamma = \frac{7}{2}\qquad\delta = \frac{13}{2}$$ and for these values $F=0$ : so, the system is consistent.
Now $$P(x)=x^4+ax^3+bx^2+cx+d=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$$ Replace $\alpha,\beta,\gamma,\delta$ by their values and make $x=\frac 12$ to get the result which is effectively $-72$.
However, please notice that the values given in your PS are wrong for $\alpha$ and $\beta$.