C- denotes continuous functions on that particular interval
P- denotes polynomial functions from the reals to the reals
From the above, I need to determine if there exist an isomorphism between V and W for each of the cases. For the first case Im convinced no such isomorphism exists for the first two and the second last unless there is something deep ive missed. For the third case, I can see a possible isomorphism as the functions in W span R^2 but not convinced there exists a bijection from a point in R^2 due to several solutions passing the a particular point. For the last case, I've an isomorphism I'm satisfied with.
For the first, consider the linear map $\tau(t) = -1+2t$. Then $\tau$ is a bijection from $[0,1]$ to $[-1,1]$. If $v \in V$, then $v \circ \tau \in W$, and similarly if $w \in W$, then $w \circ \tau^{-1} \in V$. Hence $\Phi_1: V \to W$ defined by $\Phi_1(v) = v \circ \tau$ is an isomorphism.
For the second, consider $\Phi_2: V \to W$, $\Phi_2(v)(t) = \int_0^t v(\tau) d \tau$. We note that $\Phi_2^{-1}(w)(t) = w'(t)$.
For the third, consider $\Phi_3: V \to W$, $\phi_3(v)(t) = v_1 \cos t + v_2 \sin t$. We note that $\Phi_3^{-1}(w)(t) = (w(0), w'(0))^T$.
For the fourth, note that $\dim \mathbb{R}^4 = 4$, and $\dim C[0,1] = \infty$, hence no isomorphism can exist. To demonstrate, suppose $\Phi_4: V \to W$ is an isomorphism. Note that $\{ t \mapsto t^k \}_{k=0}^\infty\subset C[0,1]$ is linearly independent, and let $v_k = \Phi_4^{-1} ( t \mapsto t^k )$. However, since $\dim \mathbb{R}^4 = 4$, there must exist $\beta \neq 0$ so that $\sum_{k=1}^5 \beta_k v_k = 0$. However, this would imply that $\Phi_4(\sum_{k=1}^5 \beta_k v_k) = \sum_{k=1}^5 \beta_k \Phi_4(v_k) = 0$, which is a contradiction.
For the fifth, note that $\mathbb{P}$ has a countable (Hamel) basis ($\{ t \mapsto t^k \}_{k=0}^\infty$), but $\mathbb{R}^\mathbb{N}$ has no countable basis. To see why $\mathbb{R}^\mathbb{N}$ has no countable basis, note that $l_\infty \subset \mathbb{R}^\mathbb{N}$ and $l_\infty$ is not separable. If $\mathbb{R}^\mathbb{N}$ had a countable basis, then since $\mathbb{Q}^k$ is dense in $\mathbb{R}^k$, it would follow that $l_\infty$ was separable. Since any isomorphism would map a basis into a basis, it follows that no such isomorphism exists.
Addendum: To show why no isomorphism can exist for the fifth:
Suppose such an isomorphism exists, call it $\Phi_5$. Note that ${\cal P}(\mathbb{N}) = \{A \}_{A \subset \mathbb{N}}$ is uncountable, and let $B=\{1_A\}_{A \in {\cal P}(\mathbb{N})}$. Note that $B \subset l_\infty \subset \mathbb{R}^\mathbb{N}$, and $\|1_A\|_\infty \le 1$, and $\|1_A-1_{A'}\|_\infty = 1$ whenever $A \neq A'$.
Now let $R = \Phi_5^{-1} B \subset \mathbb{P}$. Let $\mathbb{P}^n = \{ p \in \mathbb{P}| \partial P \le n \}$, and note that $\mathbb{P} = \cup_n \mathbb{P}^n$. Since $R$ is uncountable, the set $R_n = R \cap \mathbb{P}^n$ must be uncountable for some $n$.
Note that $\mathbb{P}^n$ is finite dimensional, and let $S$ be the subspace $S = \mathbb{P}^n \cap \Phi_5^{-1} l_\infty $. $S$ is also finite dimensional, and $R_n \subset S$. Define a norm on $S$ by $\|p\|_S = \|\Phi_5 p\|_\infty$. Note that for all $p \in R_n$, we have $\|p\|_S \le 1$, and for all $q \in R_n$ such that $q \neq p$, we have $\|p-q\|_S = 1$.
However, since the unit ball in $S$ is compact, any sequence $r_n \in B$ must have a subsequence $r_{n_k}$ that converges to an accumulation point $\hat{r}$, and since the subsequence must be Cauchy, we must have $\|r_{n_k}-r_{n_l}\|_S < \frac{1}{2}$ for $k,l$ sufficiently large, which is a contradiction.