Take $I=[-1,1]$ for simplicity and define $f:S^n\times I\to S^{n+1}$ by
$$f(x_1,\ldots, x_n, z)=(rx_1,\ldots,rx_n, z)$$
$$r=1-z^2$$
So I've picked $r$ in such a way that $\lVert f(x_1,\ldots,x_n, z)\rVert=1$. Also you can easily check that
$$f(X, z)=f(X',z')\mbox{ if and only if }(X=X'\mbox{ and } z=z')\mbox{ or }z=z'\in\{-1, 1\}$$
Now this function obviously yields a continous bijection for a normal suspension $SS^{n}\to S^{n+1}$. And thus it is a homoemoprhism (since the domain is compact).
In order to generate a continous bijection for the reduced suspension one more step has to be done. We have to collapse all elements along the arc of the sphere joining two polar poles. So let $(a_1, \ldots, a_n)\in S^n$ be a fixed point. Pick
$$g:S^{n+1}\to S^{n+1}$$
in such a way that $g(x_1,\ldots, x_n, z)=g(y_1,\ldots, y_n, z')$ if and only if ($x_i=y_i$ and $z=z'$) or ($x_i=y_i=a_i$). The existence of such map does not seem to be trivial or at least I'm not sure how to prove it yet. Perhaps it can be filled by someone else.
Anyway $g\circ f$ now induces a continous bijection $\sum S^n\to S^{n+1}$ which is a homeomorphism since the domain is compact.
For the second one try something simplier first.
Lemma. Let
$$f:X\to Y$$
be a continous map. Consider 2 relations: $\sim_X$ on $X$ and $\sim_Y$ on $Y$. Now assume that for all $x, y\in X$ we have: if $x\sim_X y$ then $f(x)\sim_Y f(y)$. Then we have a well defined continous function
$$F:X/\sim_X\to Y/\sim_Y$$
$$F([x]_{\sim_X})=[f(x)]_{\sim_Y}$$
Proof. $F$ is well defined directly from the assumption that $f$ maps related elements to related elements. We will show that $F$ is continous. Indeed, let $U$ be open in $Y/\sim_Y$. Consider two projections
$$\pi_X: X\to X/\sim_X$$
$$\pi_Y: Y\to Y/\sim_Y$$
By definition $\pi_Y^{-1}(U)$ is open in $Y$. And thus $f^{-1}(\pi_Y^{-1}(U))$ is open in $X$. To complete the proof we only need to show that
$$f^{-1}(\pi_Y^{-1}(U))=\pi_X^{-1}(F^{-1}(U))$$
because by definition of quotient map this will mean that $F^{-1}(U)$ is open. But this is trivial since for any $x\in X$ we have
$$\pi_Y(f(x))=[f(x)]_{\sim_Y}=F([x]_{\sim_X})=F(\pi_X(x))$$
$\Box$
Back to your task. Consider $g\times id$. Note that in order to talk about reduced suspension you need to pick fixed points in $X$ and $Y$. These will be $x_0\in X$ and $g(x_0)\in Y$. Pick $(x,y)\in X\times\{0\}\cup X\times\{1\}\cup\{x_0\}\times I$. It is clear that
$$(g\times id)(x, y)=(g(x), y)\in Y\times\{0\}\cup Y\times\{1\}\cup \{g(x_0)\}\times I$$
Thus the induced map is well defined and continous.
Pick any surjective map $f:S^1\to S^2$ (space filling curve). By reduced suspension you have a surjective map $f^2:S^2\to S^3$ (indeed, suspension takes surjective maps to surjective maps). You can continue this construction to get any surjective map $f^n:S^n\to S^{n+1}$. Thus for any $n > m$ you get a surjective map $f^{n, m}:S^m\to S^n$ being a composition of those defined earlier.
The other way around is analogous. First define surjective map $f:S^2\to S^1$ by collapsing $S^2$ to a ball $B^2$ (by projection), then projecting $B^2$ to $I$ (since $I=B^1$) and finally wrapping $I$ on $S^1$ (by $(\cos(x), \sin(x))$ mapping). Each of these transformations is surjective and so is their composition. By suspension and composition you get a surjective map $f^{n, m}:S^m\to S^n$ where $n < m$.
