Golf Me An OOP!

CJam, 59 bytes

q_'.e=\N/{)'?=\S/>_,(%}%/(__,*{(2$z~@f=.*\m*|}/ff{1$e=>:=N}

This finishes instantly for both test cases.

It either prints the second name name of the question or 1 (both truthy), or 0 (falsy).

Try it online in the CJam interpreter.

Idea

Because of the distinction between classes and members, the challenge boils down to creating a preorder for which the input provides a partial definition.

We define that xy iff x is a y or x has a y.

For the first test case, the input states that BA, CB and Afoo. Because of transitivity, we also have Bfoo, CA and Afoo. Also, because of reflexivity, xx is always true.

For a given input, we thus can extract the partial definition of ≺ from the statements, apply transitivity a sufficient amount of times to complete the definition of ≺ and finally answer the questions.

Code

q_     e# Push all input from STDIN and a copy.
'.e=   e# Count the number of dots/statements (C).
\N/    e# Split the original input at linefeeds.
{      e# For each line:
  )'?= e#   Pop the last character and check if it is a question mark.
       e#   Pushes 1 for '?', 0 for '.'.
  \S/  e#   Split the modified line at spaces.
  >    e#   Remove the first chunk ("Does" or "Is") for questions.
  _,(% e#   Keep the first and last element of the resulting array.
}%/    e# Split the line array into chunks of length C.
(_     e# Extract the first chunk (statements) and push a copy.
       e# The original becomes an accumulator for ≺.
_,*    e# Repeat the statements C times.
{      e# For each of the repeated statements:
  (    e#   Shift out the first name.
       e#     ["w" "x"] -> ["x"] "w"
  2$z~ e#   Copy the accumulator, zip it and dump.
       e#     [["x" "y"] ["z" "w"]] -> ["x" "z"] ["y" "w"]
  @f=  e#   Rotate the shifted out name on top and check for equality.
       e#     ["y" "w"] "w" -> [0 1]
  .*   e#   Vectorized string repetition.
       e#     ["x" "z"] [0 1] -> ["" "z"]
  \m*  e#   Swap the result with the shifted array and apply Cartesian product.
       e#     ["" "z"] ["x"] -> [["" "x"] ["z" "x"]]
       e#   This accounts for transitivity; we had ["w" "x"] and ["z" "w"],
       e#   so now we have ["z" "x"].
  |    e#   Perform set union with the accumulator to add the new pairs.
}/     e#
ff{    e# For each of the questions on the bottom of the stack.
  1$e= e#   Count the occurrences of the question pair in the accumulator.
  >    e#   Remove 0 or 1 elements from the question pair.
  :=   e#   Check for equality.
       e#   If the question pair occurs in the accumulator, this pushes the
       e#   second name of the question pair. Otherwise, it pushes 1 if the
       e#   names are equal (to account for reflexivity) and 0 otherwise.
  N    e#   Push a linefeed.
}      e#

Python 3, 431 331 308 bytes

o={}
f={}
def h(z,f):
 if z not in o:f[z]=[z];o[z]=[]
while 1:
 g=input().split(' ');r=2;l=g[-1][:-1]
 if'.'in g[3]:
  if'i'in g[1]:h(g[0],f);h(l,f);f[g[0]]+=f[l]
  if'h'in g[1]:o[g[0]]+=l,
 else:
  if'I'in g[0]:r=any(l in z for z in f[g[1]])
  if'D'in g[0]:r=any(l in o[z] for z in f[g[1]])
 if r<2:print(r)

This is the full version with comments

objects = {}
synonyms = {}

def createObject(name):
    """
    Create a object with `name` if is does not yet exist and start a synonym tree.
    """
    if name not in objects:
        synonyms[name] = [name]
        objects[name] = []

# use this to read from a file
# with open("questions.txt") as file: 
#     for l in file:
        # print(">>> " + l, end='')
        # inArg = l.replace("\n","").split(" ")


while True: # to read from a file comment this
        inArg = input(">>> ").split(" ") # and this out

        out = -1

        if '.' in inArg[3]: # statement
            last = inArg[3].replace('.','')

            if 'i' in inArg[1]: # is a
                createObject(inArg[0])
                createObject(last)
                synonyms[inArg[0]] += synonyms[last]

            if 'h' in inArg[1]: # has a
                objects[inArg[0]] += [last]

        else:# question
            last = inArg[-1].replace('?','')

            createObject(inArg[1])
            if 'I'in inArg[0]: # Is a
                out = any([last in syn for syn in synonyms[inArg[1]]])

            if 'D'in inArg[0]: # Does have a
                out = any(last in objects[syn] for syn in synonyms[inArg[1]])

        if out != -1:
            print(out)

Output for test case #1:

True
True
False

Case #2:

True
True
False
False
True

I removed the debug commands for clarity in the main program, but if you would like to see them just look in the history


JavaScript, 265 263 bytes

for(o={};i=prompt().split(/\W/);)a=i[0],b=i[1],d=i[2],b=="is"?((o[a]=o[a]||{p:[],k:{}}).p.push(d),o[d]=o[d]||{p:[],k:{}}):b=="has"?o[a].k[d]=1:alert(o[b]&&(a>"E"?b==d|(c=n=>~(p=o[n].p).indexOf(d)|p.some(c))(b):(c=n=>o[n].k.hasOwnProperty(i[4])|o[n].p.some(c))(b)))

Enter a blank string to quit.

Explanation

for(
  o={};                               // o = all objects
  i=prompt().split(/\W/);             // i = command as an array of words
)
  a=i[0],                             // a = first word
  b=i[1],                             // b = second word
  //c=i[2],                           // c = third word
  d=i[3],                             // b = fourth word
  //e=i[4],                           // e = fifth word

  // Case: <name> is a <name>.
  b=="is"?(
    (o[a]=o[a]||{p:[],k:{}})          // create the object if it does not exist
      .p.push(d),                     // add the parent to the object's list of parents
    o[d]=o[d]||{p:[],k:{}}            // create the parent if it does not exist
  ):

  // Case: <name> has a <name>.
  b=="has"?
    o[a].k[d]=1                       // set the specified property

  :
  alert(                              // display the responses to the questions
    o[b]                              // false if the queried object does not exist
    &&(

      // Case: Is <name> a <name>?
      a>"E"?                          // "Is" > "E" and "Does" < "E"
        b==d                          // check if it is itself
        |(c=n=>
          ~(p=o[n].p)                 // p = direct parents of current object
            .indexOf(d)               // check direct parents for the object
          |p.some(c)                  // check the grandparents
        )(b)

      // Case: Does <name> have a <name>?
      :
        (c=n=>
          o[n].k.hasOwnProperty(i[4]) // check if this object has the property
          |o[n].p.some(c)             // check it's parents for the property also
        )(b)
    )
  )