1. No, 10 electrons transferred is correct; 10 electrons lost by the 10 iodide ions and 5 electrons gained by each chlorine atom in the chlorate ions, for a total of 10 electrons gained. You should not add add the 10 electrons lost and gained to get 20, because they are the same 10 electrons transferred total. If you give me 10 cents, you have lost 10 cents and I have gained 10 cents, but the total number of cents transferred from you to me is 10 cents, not 20 cents. Hope this makes "sense." 
2. None of the answers correspond to a zero-order reaction (for which [A] vs. t is linear).
For a first-order reaction, ln[A] vs. t is linear. For a second-order reaction, 1/[A] vs. t is linear.
For a zero-order reaction, t1/2 = [A]o / 2k
For a first-order reaction, t1/2 = ln(2) / k
For a second-order reaction, t1/2 = 1 / k[A]o
The rate of a zero-order reaction is time-independent, since Rate = k[A]0 for a zero-order reaction (with no [A] term since [A]0 = 1).
The rate of a first-order and second-order reaction will decrease with time, because [A] decreases with time and the rate law for each has [A] in it.
So, the zero-order case can be eliminated from the possible choices. Now, which of the other two cases, first and second-order, has a half-life that depends on [A]o? Then you will know which order the reaction is, and can then tell which answer applies. 
3. For this question, you need to write down some equations that relate ΔG°, ΔH°, ΔS°, Keq, E°cell, and T.
1) ΔG° = ΔH° �?nbsp; TΔS°
2) ΔG° = –RT ln (Keq)
3) ΔG° = –nFE°cell
The question says ΔH° and ΔS° are positive.
Let's take an easy choice first: the sign of ΔG. Look at equation (1). If ΔH° and ΔS° are both positive, ΔG° can be either positive or negative, depending on how large the temperature term is. Therefore, ΔG° will not always be greater than zero at all temperatures. If T is large, –TΔS° will be large, and if it is greater in magnitude than ΔH°, ΔG° will be negative, which is less than zero. If T is small, –TΔS° can be smaller in magnitude than ΔH°, resulting in a positive ΔG° since ΔH° is positive. So, the choice "ΔG° > 0 for all temperatures" is not correct.
Now, how does E°cell change with temperature? We need to combine equations (2) and (3) to get a new equation that has both E°cell and T in it. It will look like this:
ΔG° = –nFE°cell = –RT ln (Keq)
So, E°cell = RT ln (Keq)
nF
We also have to also know what happens to Keq when we change the temperature, and this will depend on the sign of ΔH°.
ΔG° = –RT ln (Keq) = ΔH° �?nbsp; TΔS°
–RT ln (Keq) = ΔH° �?nbsp; TΔS°
From this, we can rearrange to,
Keq = e–ΔH°/RT X eΔS°/R
The last term is temperature-independent, so we can ignore it for this question. Now, if ΔH° is positive (as the question states), what will happen to Keq if T increases? You may have to plug in some numbers and compare in order to tell, but as T increases, Keq increases (if ΔH° is positive).
Now, back to our E°cell equation:
If we increase T, Keq will increase, which will cause E°cell to increase.
If we decrease T, Keq will decrease, which will cause E°cell to decrease.
Well, that's the jist of it! 
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