jmc

Let $P(x)=x^3+a{x}^2+bx+5.$ The remainder $R(x)$ has degree at most 1, so let $R(x)=cx+d.$

When $P(x)$ is divided by $(x-1)(x-4),$ the quotient is of the form $x+p,$ so write $$P(x)=(x+p)(x-1)(x-4)+R(x)=(x+p)(x-1)(x-4)+cx+d.$$Comparing the coefficients of $x^2,$ we get $a=p-5.$

When $P(x)$ is divided by $(x-2)(x-3),$ the quotient is of the form $x+q,$ so write $$P(x)=(x+q)(x-2)(x-3)+2R(x)=(x+q)(x-2)(x-3)+2(cx+d).$$Comparing the coefficients of $x^2,$ we get $a=q-5.$ Hence, $p=q.$

Comparing the coefficients of $x$ in both equations, we get $$\begin{array}{rl}b& =c-5p+4,\\ b& =2c-5p+6.\end{array}$$Subtracting these equations, we get $c+2=0,$ so $c=-2.$

Comparing the constant coefficients in the first equation, we get $5=4p+d.$ Therefore, $$P(5)=(5+p)(4)(1)-10+d=10+4p+d=\boxed{15}.$$