Javascript ES6, 212 * 80% * 75% * 50% = 63.6 bytes

Test Input

I'm using the following test data:

var A1=[
  5,6,7,            // => 5-7     # (a) group pages 
  2,3,              // => 2-3,5-7 # (b) must be properly sorted
  -9,-8,-7,         // => -10--8  # (c) allow negative numbers
  29,30,31,32,      // => 29-2    # (d) lower last digit implies increasing the 10s place
  9,10,11,12,       // => 9-11    # NOT 9-2
  36,37,38,39,40,41,42,43,44,45,46,47, 
                    // => 26-37   # NOT 26-7
  99,100,101,102,   // => 29-2    # NOT 99-2
  109,110,111,112,  // => 29-2    # NOT 109-12
],
// more tests, not specified in the question
A2=[
  120,124,       // => 120,124 # (e) handle single pages
],
A3=[
  135,136,135    // => 135-6   # (f) handle duplicates
];

Basic: 212 bytes

This version satisfies the requirements of the question (a) with all bonuses (c,d,e), but hangs on single pages and garbles duplicates.

F=(a)=>{var i=a.sort((a,b)=>a-b),l=a[0],u=l-1;a.concat('').map(x=>u=(x+="")-u-1?l=console.log(l+'-'+(u<0?u:(i=0,l.length-u.length||()=>{for(;l[i]==u[i];i++);}(),u.length-i-2||u[i]-l[i]-1||i++,u.slice(i))))||x:x)}

F(A1) --> -9--7 2-3 5-7 9-12 29-2 36-7 99-102 109-2

(The output above shows spaces instead of actual newlines for brevity) Single pages: 218 bytes

This slightly longer version additionally satisfies (e), and shows single pages as single numbers (e.g. 3-5, 7, 9-12).

G=(a)=>{var i=a.sort((a,b)=>a-b),l=a[0],u=l-1;a.concat('').map(x=>u=(x+="")-u-1?l=console.log(l-u?l+'-'+(u<0?u:(i=0,l.length-u.length||()=>{for(;l[i]==u[i];i++);}(),u.length-i-2||u[i]-l[i]-1||i++,u.slice(i))):l)||x:x)}

G(A1.concat(A2)) --> -9--7 2-3 5-7 9-12 29-2 36-7 99-102 109-2 120 124

Filter duplicates: 243 bytes

Another, somewhat longer version, which also satisfies (f). I suspect that .filter can be avoided and handling of duplicates handled in the main loop in some shorter way.

H=(a)=>{var i=a.sort((a,b)=>a-b),l=a[0],u=l-1;a.filter((x,i)=>a[i+1]!=x).concat('').map(x=>u=(x+="")-u-1?l=console.log(l-u?l+'-'+(u<0?u:(i=0,l.length-u.length||()=>{for(;l[i]==u[i];i++);}(),u.length-i-2||u[i]-l[i]-1||i++,u.slice(i))):l)||x:x)}

H(A1.concat(A2,A3)) --> -9--7 2-3 5-7 9-12 29-2 36-7 99-102 109-2 120 124 135-6