questions on binary bomb phase_3

Here is the code for phase_3
08048dd4 <phase_3>:
 8048dd4:      55                         push   %ebp
 8048dd5:      89 e5                      mov    %esp,%ebp
 8048dd7:      83 ec 28                   sub    $0x28,%esp
 8048dda:      8d 45 f8                   lea    0xfffffff8(%ebp),%eax
 8048ddd:      89 44 24 0c                mov    %eax,0xc(%esp)
 8048de1:      8d 45 fc                   lea    0xfffffffc(%ebp),%eax
 8048de4:      89 44 24 08                mov    %eax,0x8(%esp)
 8048de8:      c7 44 24 04 c5 9a 04       movl   $0x8049ac5,0x4(%esp)
 8048def:      08
 8048df0:      8b 45 08                   mov    0x8(%ebp),%eax
 8048df3:      89 04 24                   mov    %eax,(%esp)
 8048df6:      e8 f1 fa ff ff             call   80488ec <sscanf@plt>
 8048dfb:      83 f8 01                   cmp    $0x1,%eax
 8048dfe:      7f 05                      jg     8048e05 <phase_3+0x31>
 8048e00:      e8 89 04 00 00             call   804928e <explode_bomb>
 8048e05:      83 7d fc 07                cmpl   $0x7,0xfffffffc(%ebp)
 8048e09:      77 71                      ja     8048e7c <phase_3+0xa8>
 8048e0b:      8b 45 fc                   mov    0xfffffffc(%ebp),%eax
 8048e0e:      89 f6                      mov    %esi,%esi
 8048e10:      ff 24 85 28 99 04 08       jmp    *0x8049928(,%eax,4)
 8048e17:      b8 00 00 00 00             mov    $0x0,%eax
 8048e1c:      8d 74 26 00                lea    0x0(%esi),%esi
 8048e20:      eb 53                      jmp    8048e75 <phase_3+0xa1>
 8048e22:      b8 00 00 00 00             mov    $0x0,%eax
 8048e27:      eb 47                      jmp    8048e70 <phase_3+0x9c>
 8048e29:      b8 00 00 00 00             mov    $0x0,%eax
 8048e2e:      89 f6                      mov    %esi,%esi
 8048e30:      eb 39                      jmp    8048e6b <phase_3+0x97>
 8048e32:      b8 00 00 00 00             mov    $0x0,%eax
 8048e37:      eb 2d                      jmp    8048e66 <phase_3+0x92>
 8048e39:      b8 00 00 00 00             mov    $0x0,%eax
 8048e3e:      89 f6                      mov    %esi,%esi
 8048e40:      eb 1f                      jmp    8048e61 <phase_3+0x8d>
 8048e42:      b8 00 00 00 00             mov    $0x0,%eax
 8048e47:      eb 13                      jmp    8048e5c <phase_3+0x88>
 8048e49:      b8 5f 01 00 00             mov    $0x15f,%eax
 8048e4e:      89 f6                      mov    %esi,%esi
 8048e50:      eb 05                      jmp    8048e57 <phase_3+0x83>
 8048e52:      b8 00 00 00 00             mov    $0x0,%eax
 8048e57:      2d 3a 01 00 00             sub    $0x13a,%eax
 8048e5c:      05 c3 03 00 00             add    $0x3c3,%eax
 8048e61:      2d cb 03 00 00             sub    $0x3cb,%eax
 8048e66:      05 73 01 00 00             add    $0x173,%eax
 8048e6b:      2d 47 03 00 00             sub    $0x347,%eax
 8048e70:      05 47 03 00 00             add    $0x347,%eax
 8048e75:      2d 48 01 00 00             sub    $0x148,%eax
 8048e7a:      eb 0a                      jmp    8048e86 <phase_3+0xb2>
 8048e7c:      e8 0d 04 00 00             call   804928e <explode_bomb>
 8048e81:      b8 00 00 00 00             mov    $0x0,%eax
 8048e86:      83 7d fc 05                cmpl   $0x5,0xfffffffc(%ebp)
 8048e8a:      7f 06                      jg     8048e92 <phase_3+0xbe>
 8048e8c:      3b 45 f8                   cmp    0xfffffff8(%ebp),%eax
 8048e8f:      90                         nop    
 8048e90:      74 05                      je     8048e97 <phase_3+0xc3>
 8048e92:      e8 f7 03 00 00             call   804928e <explode_bomb>
 8048e97:      c9                         leave  
 8048e98:      c3                         ret    

When I look at the the conditionals and switch codes, it seems that i should only consider the following
 8048dda:      8d 45 f8                   lea    0xfffffff8(%ebp),%eax # get y
8048de1:      8d 45 fc                   lea    0xfffffffc(%ebp),%eax #get x
1) x > 1
2) x < = 7
3) x==y
i.e i can see that these are the only conditions needed to keep the bomb from blowing up,
so,
a) does this mean than i can take any two values which could satisfy these conditions
b) if not how can i determine the inputs
c) In what order does sscanf read the inputs from the stack, is it as x y or y x
d) would the the manulation of %eax in the jump statments really matter to the inputs
e) what is the purpose of assigning %esi to itself eg  8048e0e:      89 f6                      mov    %esi,%esi
f) I read about indirect jump, but i didn't quite understand it
  eg 8048e10:      ff 24 85 28 99 04 08       jmp    *0x8049928(,%eax,4)

thanks in advance