Missing data in truth table and commenting on missing value
K-maps really are the easiest.
Outputs f1 and f2 are X for 110, so they can be 1 or 0 in answer.  So in boolean algebra, 110 can be added or excluded as required.
$$f_1 = \bar a \bar b \bar c + \bar a b \bar c + \bar a b c  + a \bar b \bar c + abc$$
You have \$ab \bar c\$ as a don't care.  So looking at the inputs of your don't care you have a &amp; b at 1 and c at 0.  You look for other common terms in your function.  There are three \$\bar c\$, so there may be an advantage to add it.
$$f_1 = \bar a \bar b \bar c + \bar a b \bar c + \bar a b c  + a \bar b \bar c + abc + ab \bar c$$
Extract common factor \$\bar c\$.
$$f_1 = \bar c (\bar a \bar b  + \bar a b + a \bar b + ab) + \bar a b c  + abc$$
$$f_1 = \bar c (\bar a (\bar b  + b) + a (\bar b + b)) + \bar a b c  + abc$$
$$f_1 = \bar c (\bar a + a) + \bar a b c  + abc$$
$$f_1 = \bar c + \bar a b c  + abc$$
So 3 terms and a don't care became 1 term.
I excluded \$\bar a b c  + abc\$ from the reduction, but a similar process is required for them. Go back to origional function and see if they can be reduced.
Partial K-map:

I have solved for half of f1.  You have to finish it, but this should get you started.