What is the ph of a 0.001 62 m naoh solution?

The answer 
first of all, we should know that NaOH is a strong base. For such a product, the conentration of the OH -  is equivalent to the concentration of the NaOH itself.
that means:
[ OH -] = [ NaOH] =0.001 62 
and for a strong basis, pH can be calculated as pH = 14 + log [ OH -] 
first we compute log [ OH -] :
log [ OH -] = log (0.001 62)= -2.79

finally pH = 14 -2.79 = 11.20

1. and 2. question: in which equation? or i don't understand you, 3. a), 4. c), 5. c), 6. b), 7. d), 8. d) if small number is underlined(behind symbol) , b) if big number is underlined(in front of symbol)9. what is underlined there? 10. c)