$S(RP^2),S(CP^2)$denote suspension of real and complex projective space. Then are the first order relative homotopy group $\pi_1(S(RP^2),RP^2),\pi_1(S(CP^2),CP^2) $trivial?Why?

To define the relative homotopy groups of a pair $(X, A)$, let $i:A\to X$ be the inclusion, and write $F_i$ for its homotopy fiber. Then $$ \pi_n(X, A) = \pi_{n1}(F_i). $$ In your examples, the inclusion maps are nullhomotopic, so the homotopy fibers are $$ \Omega \Sigma \mathbb{R}P^2 \times \mathbb{R}P^2 \qquad \mathrm{and} \qquad \Omega \Sigma \mathbb{C}P^2 \times \mathbb{C}P^2, $$ respectively. Since these spaces are pathconnected, the relative homotopy "groups" in question are trivial. 

