simply u solve by k-maps, the o/p can be expressed by two ways as either sum of products or products of sum.
to do that lets examin the o/p1 for example:
o/p1 is equal one only when A=1 AND B=0 , this means that o/p1=A AND B' cause if u put any thing in this equation other than A=1 And B=0 u will get o/p=0, as u can see i expressed the o/p as sum of products by considering products as the number of times o/p will be equal to one (which is one time in our case) then get expression for each product (by putting the condition and setting A for A=1 and B' for B=0, this shoud be repeated if there is more than one product , i.e. lets say if o/p =1 also at A=0 and B=0 the another expresssion will be o/p=A'B' note products are represented by multiplication of variables and note both A and B are bared(A' and B') cause both are zero) then the last step is to put the final o/p expression as the sum of all products (in this case only AB' so o/p=AB',,, if the example was inluding A'B' then o/p=A'B'+AB')
quick representation to the second o/p2 (only one at A=B=1 the o/p2=AB)
note this can be reversed by representing by products of sums , here u consider when the o/p is equa; to zero (rather than one in the first case) and u put every variable equal to zero as not bared and every vairable equal to one as bared
quick solution to o/p1 by second methode:
o/p1 =0 at (A=0 and B=0) OR (A=1 and B=1) OR(A=0 and B=1) this really is the expression just replace each and by product and each OR by sum also each variable equal one by bared and each variable equal zero by unbared
so o/p1=(A+B)(A'+B')(A+B')
if u simplify the previous expression u should get AB'
note if u solve by k-maps u get simplified expressions from the first time
Added after 1 minutes:
Too late reply , anyway hope this helps a little bit