The symmetry of conjugacy classes of a subgroup of a finite group

3 is false. Take $G=S_6$, and let $H$ be the stabilizer of $6$, which is isomorphic to $S_5$ and has index $6$ in $G$. Let $x=(1234)$; the number of elements in the conjugacy class of $x$ in $G$ is the number of $4$-cycles in $G$, which is $90$. Of those, the ones that are in $H$ are precisely the $4$-cycles that fix $6$, which is the same number as the number of $4$-cycles in $S_5$, namely $30$ of them. So your first quotient is $$\frac{|[x_G]|}{|[x_G]\cap H|} = \frac{90}{30} = 3.$$ Now let $y=(123)$. Its conjugacy class in $G$ consists of all $3$-cycles, and there are $40$ of them; the ones in $H$ are the ones that fix $6$, and there are as many of them as there are $3$-cycles in $S_5$, namely $20$ of them. So the second quotient is $$\frac{|[y_G]|}{|[y_G]\cap H|} = \frac{40}{20} = 2.$$ Of course, neither of $2$ and $3$ divide the other.

(I think the answer for 4 is "yes", but I'm still checking) As has been shown, the answer for 4 is "no" in general.

But it is true if $H$ is normal, because in that situation the $G$-conjugacy classes of elements of $H$ split into equal-sized conjugacy classes in $H$.

To see this, assume that $a,b\in H$ are conjugate in $G$, so that there exists $g\in G$ with $gag^{-1}=b$. The number of elements in $[a]_H$ is the index of its centralizer in $H$, $$C_H(a) = \{h\in H\mid ha=ah\},$$ and the number of elements in $[b]_H$ is likewise the index in $H$ of $$C_H(b)=\{h\in H\mid hb=bh\}.$$ Now consider the action of $g$ on $H$ by conjugation: I claim that $gC_H(a)g^{-1} = C_H(b)$. Indeed, if $h\in C_H(a)$, then $ghg^{-1}\in C_H(b)$: $$(ghg^{-1})b = ghg^{-1}gag^{-1} = ghag^{-1} = gahg^{-1} = gag^{-1}(ghg^{-1}) = b(ghg^{-1}).$$ Thus, $gC_H(a)g^{-1}\leq C_H(b)$. Symmetrically, $g^{-1}C_H(b)g\leq C_H(a)$, which gives the converse inclusion. Thus, $[H:C_H(a)]=[H:C_H(b)]$, and so $|[a]_H|=|[b]_H|$, as desired.

(More generally, if $\varphi$ is an automorphism of $G$, then $\varphi(C_G(a)) = C_G(\varphi(a))$, and here $\varphi$ is the automorphism of $H$ induced by conjugation by $g$.)

4 is false. A counterexample is given with $G=\operatorname{PGL}(3)$ and with $H≅\operatorname{GL}(2)$ the subgroup of matrices of the form $$\begin{pmatrix}1&0&0\\0&a&b\\0&c&d\end{pmatrix}\text{.}$$

Choose $x$ different from $0$ and $1$. Take $$a=\begin{pmatrix}1&0&0\\0&x&0\\0&0&x\end{pmatrix}\text{.}$$ Then $a$ is in the center of $H$ so that $[a]_H = \{a\}$. But a conjugate in $G$ is $$b = \begin{pmatrix}x&0&0\\0&x&0\\0&0&1\end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1/x\end{pmatrix}$$ which is not in the center of $H$, so that $|[b]_H| > 1$.