A simple uncountable linearly independent set in $V^*$ ($V$ dual--- the space of infinite sequences of reals) is the collection of vectors:
$A^r_n = r^n$
For all nonzero real values $r$. There can be no linear relation between $M$ of these, just restricting to the first $M$ positions, because of the non-vanishing of the Vandermonde determinant. This does nothing to answer the question.
Question 4 is simple： $V^*$ is the space of all infinite sequences, and the space of finite sequences dotted into these give nontrivial linear maps. The interesting question is whether these are all of the linear maps in the absence of choice.
For question 5: if probability arguments work (i.e. if every subset of $\mathbb R$ is measurable) then $V^*$ does not have a basis.
Preliminary comment: If $m$ and $n$ are two positive integers both distributed with probability distribution $P$, their sum $m+n$ does not have the same distribution $P$. To prove this, let $N>0$ be the first integer position where $P(N)$ is nonzero. Then $m\ge N$, $n\ge N$, so $m+n\ge 2N$, so the probability that $m+n$ is $N$ is zero.
Suppose that there is a basis set $B$. Then any vector $v$ in $V^*$ has an integer degree $n$ and $n$ uniquely determined elements of $B$, $e_1,...,e_n$ such that $v$ is a linear combination of the $e$'s. Generate the sequence $A_n$ by picking a Gaussian random number with width $1$ at each position $n$. Generate $B_n$ in the same way. Then the degree of $A$ is some integer $n$, with probability distribution $P(n)$, and the degree of $B$ is an integer $m$ with a probability distribution $P(m)$. The degree of $(A+B)/\sqrt{2}$ is then $m+n$ by adding the expansion of $A$ and the expansion of $B$ (there is no chance that $A$ and $B$ share basis elements, the probability of meeting any finite dimensional subspace is zero), and its distribution is the distribution of the sum of two random integers with distribution $P$. But $(A+B)/\sqrt{2}$ is identically distributed with $A$ and $B$, therefore its degree must have distribution $P$, and there is no such distribution $P$ on positive integers.
I'm repeating the argument in the form I came up with so as to make the rest of the answer more obvious. A closer inspection reveals that the nonmeasurable sets here are the sets of all vectors with degree $n$. These union up to the whole space, so they need to add to the full measure, but the larger $n$ degrees necessarily swamp the lower $n$ degrees in measure because of the properties of linear combination, forbidding a consistent assignment of measures.
For the main question, questions 1,2,3, I show that the existence of any map $f$ in $V^{**}$, other than the ones given by dotting with elements of $V$, either leads to a probabilistic contradiction or a nonprincipal ultrafilter on the integers. Both of these cannot be "explicit", so there is no explicit linear map in $V$ double-dual other than the canonical images of those in $V$.
Proposition: $f$ of any infinite sequence of independent Gaussian random numbers is Gaussian distributed.
Proof: let two picks of the sequence be $S$ and $S'$, $(a S+ b S')\over \sqrt{a^2 + b^2}$ is identically distributed as $S$ and $S'$, so $(a f(S) + b f(S'))\over\sqrt{a^2 + b^2}$ is identically distributed as $f(S)$. Note also that $f(-S)=-f(S)$, so that the probability distribution of $f$ is symmetric around zero. This implies that $f$ has the convolution properties of a Gaussian symmetric about zero, and is therefore a Gaussian with zero mean (or a delta function at zero, which I will call a Gaussian of zero width below).
This means that the linear function $f$ gives a map $\Sigma$ from every sequence of variances to a variance. The function $f$ has variance $\Sigma(\sigma_k)$ when evaluated on the sequence of Gaussian random variables of variance $\sigma_k$
Proposition: If the variances $\sigma_k$ are each individually bigger than $\sigma'_k$, then $\Sigma(\sigma)>\Sigma(\sigma')$
Proof: Let $\alpha_k$ be such that $\alpha_k^2 + \sigma_k^{'2} = \sigma_k^2$. Generate a sequence of Gaussian random picks $A_k$ with variance $\sigma'_k$, and a second Gaussian sequence $B_k$ with variance $\alpha_k$. By construction, $A_k+B_k$ has variance sequence $\sigma_k$.
$f(A)$, $f(B)$, and $f(A+B)=f(A)+f(B)$ are each Gaussian distributed, and the variance of $f(A)$ squared plus the variance of $f(B)$ squared equals the variance of $f(A+B)$ squared. This proves the result.
Theorem: Define the sequence $A^N_n$ to be zero for $n\le N$, and a Gaussian random number with unit variance for $n>N$. If probabilistic arguments work, there is an $N$ for which $f$ on $A^N$ is zero with probability $1$.
Proof: If $f$ on $A^k$ is nonzero for all $k$, consider the infinite sum
$S = C_1 A^1 + C_2 A^2 + C_3 A^3 + C_4 A^4 + \cdots$
since each $A^k$ is zero on the first $k$ positions, this just defines $S$ as a sequence of Gaussians of increasing variance (this is not really an infinite sum--- it is just a trick for writing an infinite sequence of variances in a more illuminating way). For any sequence $C_1, C_2, C_3, \ldots$, the function $f$ produces a Gaussian with a given width which is an non-decreasing function of the $C$'s.
By assumption, $f(A^k)$ has some nonzero variance $a_k$ for all $k$. Choose $C_k$ to be $1/a_k$. Then the width of $f$ on $S$ is greater than any integer $k$, a contradiction.
Therefore there is an integer $N$ such that $f$ has zero variance acting from some $A^N$ on, i.e. $f$ acting on a sequence which is zero at the first $N$ positions, and Gaussian random with unit variance on the remaining positions, gives $0$ with certainty.
What remains is to deal with the case that $f$ is nonzero on some specific vector $v$ which has no chance of ever being generated by random Gaussian picking. To deal with this, you need the following:
Theorem: If $f$ is of zero width on all Gaussian random variables, and $f$ is nonzero on some vector $v$ (and probability works), then there is a nonprincipal ultrafilter on the integers.
proof: Suppose $f(v)$ is nonzero for some infinite sequence $v$. By the zero width property, $f$ is zero on sequences which are nonzero only at finitely many places.
Define a set $S$ of integers to be "zeroing" iff: for any collection $g(S)$ of independent Gaussian random variables defined on $S$, and a Gaussian random variable $g$, $f$ acting on $g v_S + g(S)$ is certainly zero. $g v_S$ is $g$ times the restriction of $v$ to zero on the complement of $S$, while $g(S)$ is a sequence of random variables inside $S$, and zero outside $S$.
Any finite set is zeroing, while the full-set is nonzeroing since $f(gv+g(N))$ has the variance of $g$ times $f(v)$ (since $f$ is zero acting on the Gaussian random sequence $g(N)$).
If $S$ is zeroing, $S$ complement is nonzeroing by linearity. (Note that the converse is not true--- $S$ can be nonzeroing and also $S$ complement nonzeroing--- this is not automatically an ultrafilter).
If $S$ is zeroing and $S'$ is a subset of $S$, then $S'$ is zeroing (by the positivity of adding Gaussian widths).
If $S$ is zeroing and $S'$ is zeroing then $S$ union $S'$ is zeroing, since the sum of independent Gaussian random variables on $S$ and $S'$ are again independent Gaussian random variables on $S$ union $S'$, and any independent Gaussian random variables on the union of $S$ and $S'$ can be decomposed in this way. Therefore the nonzeroing sets form a nonprincipal filter extending the finite-complement filter.
Using dependent choice, either you have an infinite sequence of disjoint restrictions of $v$ $S_1$, $S_2$, $S_3,\ldots$ which are nonzeroing, or the restriction onto one of the sets makes an ultrafilter. An infinite sequence of disjoint nonzeroing restrictions leads to a probabilistic contradiction, as before, by considering
$A = \sum_k C_k g(S_k)$
with appropriate fast-growing choice of $C_k$, just like before. The terminating case gives an ultrafilter.
Putting the two theorems together, the function $f$ has to be certainly zero on Gaussian sequences from position $N$ onward, so $f$ is a linear function on the first $N$ values plus a residual. The residual is zero on the first $N$ positions, and makes a linear function on the values past $N$, and the residual is zero on any Gaussian random sequence.
If the residual function is nonvanishing on some vector $v$, then it either leads to a probabilistic contradiction or it defines a nonprincipal ultrafilter on $\mathbb Z$. So in a universe with every subset of $\mathbb R$ measurable, with dependent choice, and with no nonprincipal ultrafilter on the integers, $V$ double dual is the canonical image of $V$, and questions 1,2,3 are answered.
