a.
[tex]\dfrac{\mathrm dP}{\mathrm dt}=0.05P-0.00005P^2=0.00005P(1000-P)[/tex]
so that [tex]\frac{\mathrm dP}{\mathrm dt}=0[/tex] for [tex]P=0[/tex] or [tex]P=1000[/tex]. The derivative being equal to 0 gives you some information about the long-term stability of the population [tex]P[/tex]; namely, that the carrying capacity is 1000.
b. If [tex]0<P<1000[/tex], then [tex]0.00005P>0[/tex] and [tex]1000-P>0[/tex], so that [tex]\frac{\mathrm dP}{\mathrm dt}>0[/tex], which means the population would be increasing.
c. If [tex]P>1000[/tex], then [tex]0.00005P>0[/tex] but [tex]1000-P<0[/tex], so [tex]\frac{\mathrm dP}{\mathrm dt}<0[/tex] and the population would be decreasing.