1. Since $E_1,E_2,E_3$ are events, their probabilities must satisfy $0\le P(E_i)\le 1$ for each $i$.

2. Given: $P(E_1)=\frac{2+3p}{6},\qquad P(E_2)=\frac{2-p}{8},\qquad P(E_3)=\frac{1-p}{2}.$

3. Use non-negativity of each probability.

   From $P(E_1)\ge 0$: $\frac{2+3p}{6}\ge 0 \implies 2+3p\ge 0 \implies p\ge -\frac{2}{3}.$

   From $P(E_2)\ge 0$: $\frac{2-p}{8}\ge 0 \implies 2-p\ge 0 \implies p\le 2.$

   From $P(E_3)\ge 0$: $\frac{1-p}{2}\ge 0 \implies 1-p\ge 0 \implies p\le 1.$

   Combining these: $-\frac{2}{3}\le p\le 1.$

4. Now check the upper bound $P(E_i)\le 1$.

   For $E_1$: $\frac{2+3p}{6}\le 1 \implies 2+3p\le 6 \implies p\le \frac{4}{3}.$

   For $E_2$: $\frac{2-p}{8}\le 1 \implies 2-p\le 8 \implies p\ge -6.$

   For $E_3$: $\frac{1-p}{2}\le 1 \implies 1-p\le 2 \implies p\ge -1.$

   These are weaker than the earlier bounds, so still: $-\frac{2}{3}\le p\le 1.$

5. Since the events are mutually exclusive, we must also have $P(E_1)+P(E_2)+P(E_3)\le 1.$

   Compute the sum: $\frac{2+3p}{6}+\frac{2-p}{8}+\frac{1-p}{2}.$

   Taking LCM $24$: $=\frac{4(2+3p)+3(2-p)+12(1-p)}{24}$ $=\frac{8+12p+6-3p+12-12p}{24}$ $=\frac{26-3p}{24}.$

   Therefore, $\frac{26-3p}{24}\le 1$ $26-3p\le 24$ $-3p\le -2$ $p\ge \frac{2}{3}.$

6. Combine with the earlier range: $\frac{2}{3}\le p\le 1.$

   Hence, $p_2=\frac{2}{3},\qquad p_1=1.$

7. Therefore, $p_1+p_2=1+\frac{2}{3}=\frac{5}{3}.$

So the correct option is B.