a)Lemma : let $P$ be an arbitrary point in plan,and $G$ be the centroid of triangle $ABC$ then We have: $PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3PG^2$ proof.let $D$ and $A'$ be midpoints of $AG$ and $BC$ use stuart theorem in triangles $AGP,DPA',PBC$ and do same for $B$and $C$.do the rest of ir yourself. so now your problem: in equilateral triangle(let side lenght be $a$) we have $O\equiv I\equiv G\equiv H\equiv \dots$ so from lemma we have: $PA^2+PB^2+PC^2=3(\frac{\sqrt 3}{3})^2a^2+3(\frac{\sqrt{3}}{6})^2a^2=\frac 54a^2$for $a=2$ it will be $5$

Hey Mathx, can the 2nd part be solved by the similar idea as the one you have used in the 1st part????

Yes the 2nd part can also be solved using the same idea. Moderators, please move this to the Geometry Solved Problems.