1) There are a few possible solutions. What's important to understand that 
instances exist independently of orders and that instances are then assigned to orders and shipments. 

orderdest(orderid, shipto) 
    orderid is the key; shipto is destination loc.
orderitem(orderid, item, quantity)
    (orderid, item) is key 
location(loc, address)
    loc is key
shipment_loc(shipid, source_loc, dest_loc)
    shipid is key 
	[shipid, source_loc could also be the key; ]
ordership(orderid, shipid) -- both together are key
    [It's conceivable to combine ordership and shipment.]
locitem(item, instanceid, loc)
	item and instanceid (or conceivably just instanceid) is the key 
shipment(shipid, item, instanceid)
	shipid and instanceid constitute the key 
	[It's also possible to put these tables together]
orderinstance(orderid, item, istanceid)
    orderid, item, istanceid constitue the key (or conceivably just orderid, instanceid)

2)

-- For a given orderid O. Select all items that are in a shipment and their source location

select item, instanceid, address from location L, shipment S, shipment_loc SL, ordership OS 
    where 
    S.shipid = OS.shipid  and 
    S.shipid = SL.shipid  and 
    S.source_loc = L.loc and 
    OS.orderid = @O

-- For a given orderid O. Select all items that are in a certain location (haven't been
shipped yet)

select item, instanceid, address from location L, locitem LI. orderinstance OI where 
    LI.loc = L.loc and 
    OI.orderid = @O and 
    OI.instanceid = LI.instanceid

3)

--- The orderid used in the query is: O ship to location: X. As we want to get all the instances with the
maximum distance we first create a temporary table containing the distances of the order. Also possible to
get the shipto location from orderdest.

create temporary table temp select D.dist from  locitem LI, distance D, orderinstance OI  
    where 
    OI.orderid = @O and 
    OI.instanceid = LI.instancedid
    D.loc1 = I.loc and 
    D.loc2 = @X;

--- Now select all instances with distance equal to max
select item, instanceid from  locitem LI, distance D, orderinstance OI  
    where 
    OI.orderid = @O and 
    OI.instanceid = LI.instancedid
    D.loc1 = I.loc and 
    D.loc2 = @X and
    D.dist = (select max(dist) from temp);
    
It is also possible to perform the above query without a temporary table and just with a more complex
sub-select statement.


4)
a) The set is already minimal.

b) AB is a key. BE is also a key.

c) Combining the FDs with union rule we get: 
A  -> CE
B  -> C
CE -> DA
DE -> C

We can now create the tables: (B,C), (C, E, D, A) which contain all relationships. But need to 
add a superkey. So we get:

(A, B), (B,C), (C, E, D, A)
An additional decomposition is:
(B,E) (B,C), (C, E, D, A)
Two other alternatives:
(A, B), (B,C), (C, E, D), (C, E, A)
(B, E), (B,C), (C, E, D), (C, E, A)


d)
A -> C, B -> C, DE -> C, 
C -> E, C -> D, C -> A

Another answer:
A -> C, B -> C, A->E, 
C->D, DE->C, C->A




5) 

An example is:

A --> B, AB --> C, C --> B, 

Step 1 can't remove any FDs.
Step 2 can remove: B from AB-->C (as A-->B). 
Thus we get: A --> B, A --> C, C --> B, 
And now Step 1 can remove A --> B

6) 

The most basic decompisition is at least a separation of the global properties characterizing the room
from the reservation itself. Thus a basic decompistion would be:

inventory(hotelid, roomtypeid, date,  numbertaken) - (hotelid, roomtypeid, date) is a key
roomtype(hotelid, roomtypeid, numberavailable, description) - (hotelid, roomtypeid) is a key.

If roomtypeid is defined as globaly unique an alternative solution is:

inventory(roomtypeid, date,  numbertaken) - (roomtypeid, date) is a key
roomtype(roomtypeid, hotelid, numberavailable, description) - (roomtypeid) is a key.


7) 

<answer>
{
for $i in doc("hispo.xml")//item
return <item>{$i/@partNum}<totalSales>{$i/USPrice/text() * $i/quantity/text()}</totalSales></item>
}
</answer>


Here is another version with variables:

<answer> 
{
for $i in doc("hispo.xml")//item
let $p := $i/USPrice/text()
let $q := $i/quantity/text()
let $n := $i/@partNum

return:
<item> {$n}
<totalSales> { $p*$q } </totalSales>
</item>
} 
</answer>

The result is:

<answer>
  <item partNum="872-AA"><totalSales>148.95</totalSales></item>
  <item partNum="926-AA"><totalSales>79.96</totalSales></item>
</answer>