Answer
$C \approx36.2^\circ$, $A= 43.8^\circ$, $a\approx3.51$. Only one triangle is possible.
Work Step by Step
Step 1. Based on the given conditions, use the Law of Sines, we have $\frac{sinC}{3}=\frac{sin100^\circ}{5}$, thus $C=sin^{-1}(\frac{3sin100^\circ}{5})\approx36.2^\circ$ or $180-36.2=143.8^\circ$ (discard this one as B+C>180)
Step 2. We have $A=180-100-36.2=43.8^\circ$
Step 3. We have $\frac{a}{sin43.8^\circ}=\frac{5}{sin100^\circ}$, and $a=\frac{5sin43.8^\circ}{sin100^\circ}\approx3.51$. Only one triangle is possible.