I didn't ignore it - yet it didn't help. The instructions are ambiguous. How are you supposed to fill them in? What is the connection between the 8 in the first example and 8 counters? Are they the same by accident? Why should both results be the same (8+3 and 10+X) - I thought about adding 10 new counters, so that's 8+3=11, 10+8=18, 8+10=18 (although it doesn't make much sense either). Then thought about adding 3 to the lower box (not good). There are simply not enough information to "solve" this. Even if most people can guess what the right solution is, I don't think that's good enough for homework.
In general, homework is not intended to contain the entire lesson all over again. That means parents often don't know what to do, or how to help.
Further, lessons and homework are often not intended to be done by creative people without instruction. Creative people find unexpected ways to follow instructions the teachers (or lesson planners) thought were clear, obvious and unambiguous.
I see a lot of criticisms of lessons and homework like this, but I've also seen the frustration from teachers trying to create lessons that are EXCITING and ENGAGING and ENTERTAINING and THRILLING and WONDERFUL, because now they're not expected just to be good at teaching and experts in their subjects, but also entertainers.
This lesson/homework isn't perfect, but I think many of the criticisms aimed at it are misplaced and ignoring the context. This homework is being assessed as if it were a lesson, and that's unfair.
Finally, homework that my colleague's child is set is accompanied by a sheet explaining the lesson. My colleague hos no trouble helping with the homework, and constantly praises the variety of approaches and integrity of the work.
My answer to the original question is, "No, but I would if I'd seen it before." The children, presumably, have seen it before.
However, this is a pretty good way of learning a general principle. Take the first example, "8+3". Over on the right you have boxes in groups of 10, and the first box has eight "counters" filled in. Now you add three by filling in another three boxes. This gives you ten full boxes, and another one left over. Thus 8+3 goes to 10+1.
ADDED IN EDIT: Now that I think of it, how many of you simply ignored the instructions to fill counters in the boxes to help you, and tried to see what was going on just by looking at the numbers?
Of the three that you are adding, two are used to make the eight up to ten, so you only have one left. In a sense you've moved 2 from the 3 into the 8, making the sum easier. Moving stuff around in an addition is very similar to moving stuff around (and then making a correction) in multiplication. This is how Art Benjamin squares two and three digit numbers faster than you can punch them into a calculator.
53 squared is 53 times 53. Move three from one into the other (remember the 3 you've moved around) so you get 50 times 56. Use the same trick but in a different way: move a factor of two from the 56 into the 50 so you get 100 times 28, or 2800. Now add on the square of the three you moved earlier, giving 2809.
ADDED IN EDIT: How this works can be demonstrated clearly with an appropriate diagram.
If you're multiplying, moving factors is the same sort of operation (both multiplications) so you don't need a correction. If you move by addition/subtraction it's a different operation, so you will need a correction.
However, all of this fits into a larger framework. Getting only a small part lets you do those bits, but as it fits together you end up with more than just the parts, you get a much larger framework.
This is actually, tangentially, related to the item I submitted 10 hours ago and which sank without a trace:
http://news.ycombinator.com/item?id=951250
The same trick is used twice in two different context to give the Infinite Ramsey Theorem: Every infinite graph contains an infinite complete subgraph or an infinite null graph.
Very similar.