Stronger two-space formulation of Hurewicz theorem about homotopy and homology groups The following theorem of Hurewicz holds (let $\cdot$ be the one-point space and $n\!\geq\!2$): 
If $\pi_i(X)\cong\pi_i(\cdot)$ for $i\!<\!n$, then $H_i(X)\cong H_i(\cdot)$ for $i\!<\!n$ and $\pi_n(X)\cong H_n(X)$.
Does the following hold (I suspect not): 
(1) If $\pi_i(X)\cong\pi_i(Y)$ for $i\!<\!n$, then $H_i(X)\cong H_i(Y)$ for $i\!<\!n$, and $\pi_n(X)\cong H_n(X)$?
Does the following hold: 
(2) If $\pi_i(X)\cong\pi_i(Y)$ for all $i$, then $H_i(X)\cong H_i(Y)$ for all $i$?
Does the following hold: 
(3) If $H_i(X)\cong H_i(Y)$ for all $i$, then $\pi_i(X)\cong \pi_i(Y)$ for all $i$?
If yes, how can I prove them, preferrably using the Hurewicz theorem above. If not, are some additional mild hypotheses sufficient to make it work?
Do (2') and (3') hold, in which $H_i$ has been replaced by $H^i$?
 A: (2) is true if the isomorphisms on $\pi_n$ are induced by a map $f\colon X\to Y$. This follows from Whitehead's theorem. But in general the answer is no. Wikipedia's article on the Whitehead theorem gives the example of $X=S^2\times \mathbb{RP}^3$ and $Y=\mathbb{RP}^2\times S^3$. They have isomorphic homotopy groups, since they have the same fundamental group $\mathbb Z_2$, and have the same universal covers. But they have different homology groups, which you can see by using the Künneth formula.
A: (1) does not hold. For the last part ($\pi_n(X) \simeq H_n(X)$), simply take $X=Y$ and find a counterexample to Hurewicz's theorem if you don't assume $\pi_k(X) = 1 \forall k < n$. For the first part, take the following counterexample (with a big enough $n$ st. $H_n(X) \not\simeq H_n(Y)$):
(2) does not hold either. For example, $X = S^2 \times \mathbb R P^3$ and $Y = S^3 \times \mathbb R P^2$ have the same homotopy groups, but not the same homology (see Wikipedia's article and Grumpy Parsnip's answer). The additional condition needed is that the isomorphisms $\pi_*(X) \simeq \pi_*(Y)$ must be induced by a map $f : X \to Y$.
(3) does not hold either. $S^2 \times S^4$ and $\mathbb C P^3$ have the same homology groups ($\mathbb Z$ in degrees 0,2,4,6 and zero elsewhere), but different homotopy groups ($\pi_2(\mathbb C P^3) = \mathbb Z$ but $\pi_2(S^2 \times S^2) = \mathbb Z^2$).
For (1) and (3) I don't know of nice additional conditions that would make the theorems true. Note that the counterexamples to (2) and (3) also hold in cohomology (use the universal coefficient theorem).
