You should think of this as a two-part problem: first you must pick the $7$ letters to be used, and then you must arrange them. You can’t arrange them until you’ve picked them. Picking them is just picking sets of things: order is irrelevant, so you’re counting combinations. Arranging them, on the other hand, is clearly a matter os specifying an order, so you’re dealing with permutations.
Are there other ways to solve the problem? Yes, but they’re more difficult. You could begin by picking an ordered string of $4$ consonants; this can be done, as you said, in $_7P_4$ ways. You now have a skeleton $s_1C_1s_2C_2s_3C_3s_4C_4s_5$, where $C_1,C_2,C_3$, and $C_4$ are the consonants, and $s_1,s_2,s_3,s_4$, and $s_5$ are the slots into which you can insert vowels. There are now $_5P_3$ ways to select an ordered string $V_1V_2V_3$ of $3$ vowels, and the problem is to count the ways to fit these vowels into the $5$ open slots in the consonant skeleton. Doing that is a matter of selecting a multiset of $3$ not necessarily distinct slots from the $5$ available. This can be done in
$$\binom{5+3-1}3=\binom73={_7C_3}=35$$
ways, so the there are $840\cdot60\cdot35=1,764,000$ such words, exactly the figure obtained by the other computation.
For $(a)$, there are $3$ positions to choose and for each position, there are $2$ colors available. Thus the answer is $3 \cdot 2 = 6$.
For $(b)$, there are $\binom{3}{2}$ position combinations and for each position combination, you can choose $4$ color combinations (i.e., RR, RB, BR, BB), thus there are totally $3 \cdot 4 = 12$ ways.
For $(c)$, for convenience, we temporarily suppose that there are $3$ red flags and $3$ blue flags. We count the total # of ways under this assumption and then remove those cases that use $3$ flags of same color. For each position, we have three options: red, blue and empty. Therefore, the total # of ways is $3^3 = 27$, in which there are two cases where we use all red flags or all blue flags. Thus, we conclude that the answer is $27 - 2 = 25$.
Best Answer
Just a start:
For part (a): Each flag can go on any of the 10 poles. That is 10 choices for each of the 25 flags to be placed, so number of ways 10*10*...*10 with 25 copies of 10 being multiplied, i.e. $10^{25}.$
This is an application of the multiplication principle. To make that more see-able suppose at a restaurant they offer 3 entrees and 2 desserts. Then in all you have 3*2=6 ways to choose your meal. In general if an overall task T can be described as a sequence T1,T2,...,Tn of smaller tasks, then the number of ways to do T is obtained on multiplying each of #(T1),#(T2),...,#(Tn), where by #(Tk) is meant the number of ways to do task Tk.\
Parts (b),(c) will be more complicated--- I'll leave that to another answerer for now or maybe you can get it.