Step 1:

Given probability of first examination P(I)=0.8

P(II)=0.7

Probability of atleast (I or II)=$P(I \cup II)=0.95$

$\therefore P(I \cup II)=P(I)+P(II)-P(I \cap II)$

$0.95=0.8+0.7-P(I \cap II)$

Step 2:

$\therefore P(I \cap II)=1.5-0.95$

$\Rightarrow 0.55$

$\therefore$ Probability of passing both is 0.55

Hence (A) is the correct answer.