I'm not entirely sure what you mean by "First in a baby," so I'm just going to assume that you mean you have two fair spinners, the first having an equally likely chance to land on 1, 2, 3, or 4. The second has equally likely chances to land on 5, 6, 7, or 8. One way to look at this problem is just to figure out what each combination would add up to, and then go from there. We know that there are a total of 16 options, 4 for the first spinner times 4 for the second. So, starting with one on our first spinner, we have
1+5=6
1+6=7
1+7=8
1+8=9
So right away, we know that if we get a 1, we aren't going to get 10. Since we have an equally likely chance of getting any of 4 options, we have a 4/16 chance of getting a 1, so that means at most we can get a 10+ 12/16 of the time. Now let's see what happens if we roll a 2.
2+5=7
2+6=8
2+7=9
2+8=10
Here, we have one option, 2+8, where we get a 10 or higher. The other 3, we don't. So, as there's a 4/16 chance of rolling a 2, in 1/16 so far, we get a 10 or higher. Now let's see if we get a 3 first.
3+5=8
3+6=9
3+7=10
3+8=11
Here we have two options with a 10 or higher, so now we have 3/16 cases in which we get 10 or higher, 1 from before, two more now. Finally, what happens if we get a 4 first?
4+5=9
4+6=10
4+7=11
4+8=12
There are 3 options here, with 3 before, so in total, we have 6/16 scenarios where we get a 10 or higher. Or to simplify, 3/8.