First we are going to make some obvious restriction:
$$m \le k \le n$$
Note that if we choose the first number then the following $m-1$ numbers can't be chosen. For example in your example if we choose $2$ to be our first number, then it implies that $3,4,5,6$ are the second, third, fourth and fifth number respectivly.
There are $n-m+1$ to choose the first number, because if we choose a number greater that $n-m$ then we can't choose $m$ consecutive number starting with that one.
Now it's we need to chose $n-m$ random numbers for $k-m$ places.
Because you've mentioned lottery tickets, where the order isn't important we'll assume that in the calculation the order isn't important.
First I want you to note something. I'll again use your example. We take numbers from 1-5 as consecutive and we'll place them in the first 5 places, and then some random number but for the purpose of presentation we'll choose 6. Then the set will be:
$${1,2,3,4,5,6}$$
Now let's think like these, we'll choose the numbers from 2-6 and will place them in the last 5 places and the random number will be 1. The set then will be;
$${1,2,3,4,5,6}$$
Aren't this sets the same? But we choose the using 2 different ways, so that means we'll make some restrictions and we'll make 2 cases.
Case 1: $k=m$
Obviously we have space only to place the consecutive number so let $P(n,k,m)$ represent the number of ways to choose $k$ numbers out of $n$ numbers with $m$ consecutive numbers, so we have:
$$P(n,k,m) = n-m+1$$
Case 2: $k \ge m+1$
Note that if $1$ is the first number of the series we can't choose the predecessor, because there isn't one. So we have:
$$P(n,k,m) = (n-m) \times \binom{n-m-1}{k-m} + \binom{n-m}{k-m}$$
And if we calculate for your example we have:
$$P(49,6,5) = (49-5) \times \binom{49-5-1}{6-5} + \binom{49-5}{6-5}$$
$$P(49,6,5) = 44 \times \binom{43}{1} + \binom{44}{1}$$
$$P(49,6,5) = 44 \times 43 + 44$$
$$P(49,6,5) = 1936$$
For (1), note that the objects are not all distinct: e.g. the green blocks. consider the green blocks in isolation. We can arrange them in $4!$ ways if they were distinct- however, as they are all the same colour, then we need to divide by a factor of $4!$, giving a total arrangements of 1. This is because we can swap the green blocks around with each other any way we like and still get the same arrangement.
Now, if all the blocks of the same colours are together- all the greens are together, all the blues are together and all the reds are together. The greens can be arranged amongst themselves in only 1 way, as I showed; similar logic goes for the red and blue blocks. So now we have 3 objects to arrange: the blob of green blocks, red blocks and blue blocks. They are all distinct objects, so $3!=6$ ways.
As for (2)- let's consider the digits 3, 8 and 5 in isolation. There are 3!=6 arrangements of these 3 digits. Only one arrangement has this particular order: 3,8,5. So it follows that only $\frac{1}{6}$ of the total arrangements will satisfy the given condition.
Hence, for (2) the answer is $\frac{8!}{6}$
Best Answer
There are useful general techniques that apply to problems like these.
With distinct numbers, to do this fast, we should flip the picture around with the idea of Young diagrams. Three rows of different lengths, totalling 15 boxes is the same as a bunch of columns of lengths 1,2,3 with at least one of each length. Here's an example diagram for 10 (as opposed to 15) from Wikipedia:
Then we can break things down not by smallest number, but by number of threes.
1+2+4+5=12, which is the answer if "whole numbers" means positive integers, as I think it does in this context.
And if "nonnegative integers" was meant, then we also need to include two-row Young diagrams. How can you build 15 out of 2s and 1s alone? You could have anywhere from one 2 to seven 2s, so you get 7 new things, so 12+7=19, as John found.
As an aside, if you weren't restricted to 2-3 minutes, and knew about generating functions, you could take this idea quite far.
If we didn't have the condition that the numbers were distinct, then the problem reduces to stars and bars: If the numbers $a$ $b$ and $c$ have to be positive integers, then $a-1$, $b-1$, and $c-1$ are nonnegative integers that sum to $15-3=12$, so that two bars and twelve stars should do it: ${14}\choose{2}$.