I will attempt to answer with the assumption that you do not have the answers, so you can see the thinking process.
(a) 1
The sample space should be defined as a hand of $10$ cards. I like to use the term "a vector of $10$ cards". As you mentioned, the sample space has $52^{10}$ possibilities. The final answer is $$52^{10}$$
(a) 2
The sample space is defined as a hand of $10$ cards with no Aces, $A\clubsuit \space A\spadesuit \space A\heartsuit \space A\diamondsuit $. There are $48$ different draws per card of a hand. Hence, the sample space has $48^{10}$ possibilities. The final answer is $$\frac{48^{10}}{52^{10}} = \begin{pmatrix}\frac{48}{52}\end{pmatrix}^{10}$$
(a) 3
For this question, instead of considering the sample space with $1K, 2K, 3K, 4K$...only consider the space where there are $0K$. What you can do is use the answer from (a) 2...simply swap out $A$ to $K$. The final answer is
$$1-\frac{48^{10}}{52^{10}} = 1-\begin{pmatrix}\frac{48}{52}\end{pmatrix}^{10}$$
(a) 4
To obtain the number of possibilities with $1Q$, consider this:
There are 4 ways to draw a $Q$ and 48 ways to draw any other card. The number of possibilities to draw will be
$$\begin{bmatrix}\frac{4}{52} \times \begin{pmatrix}\frac{48}{52}\end{pmatrix}^9\end{bmatrix}$$
However, there are $10$ ways to arrange these draws. Divide this by the whole sample space (with no restrictions), you get
$$\frac{\begin{bmatrix}\frac{4}{52} \times \begin{pmatrix}\frac{48}{52}\end{pmatrix}^9\end{bmatrix}}{52^{10}}$$
Your question about drawing a $4,6$ or a $6,4$ is a good one.
I'd break it down into two cases: one with the cards the same, the other with cards different.
If the cards are the same, first choose the face value of the card ($13$) then choose the two suits ($_4C_2 = 6$). So there are $13 \cdot 6 = 78$ possible ways to draw a pair. Of these, two $5$'s through two $8$'s satisfy your criterion: $4$ values $\times 6$ suit combinations for each $= 24$ possibilities.
If the cards are different, first choose the two face values ($_{13}C_2 = 78$) then choose the suit for each card ($16$). So there are $78 \cdot 16 = 1248$ ways to draw two different cards.
(Note: $_{52}C_2 = 1326 = 78 + 1248,$ so we've indeed accounted for all possibilities.)
Counting the number of possibilities with the sum between $10$ and $16$ with two different cards is bookkeeping. I'll start with the larger card, and work my way up from the bottom with the smaller cards to go through the possibilities methodically. I'll also not write a higher card second, because that would double-count.
$A2, A3, A4, A5$
$K2, K3, K4, K5, K6$
$Q2, Q3, Q4, Q5, Q6$
$J2, J3, J4, J5, J6$
$T2, T3, T4, T5, T6$
$92, 93, 94, 95, 96, 97$
$82, 83, 84, 85, 86, 87$
$73, 74, 75, 76$
$64, 65.$
I count $42$ combinations for the values. Since either card can be any suit, that's $42\cdot 16 = 672$ possible combinations.
Now we're basically done:
$$P = \frac{672+24}{1326} \approx 0.525.$$
Best Answer
Just sum the probabilities of $1-2, 2-3, 3-4,..., 12-13$
probability to get $1-2$ = $2* \frac{4}{52}*\frac{4}{51}$
The total probability = $\frac{2*12*4*4}{52*51}$