### Part (a)

To prove that if $F$ is injective (1-1), then it is surjective, we will use the properties of compactness and continuity in the product topology.

1. **Consider the Set $S$**:
- Let $S$ be the set of all infinite sequences in $\{0,1\}^{\mathbb{Z}}$ such that $a_i = 0$ for $i \leq 0$ or $i > N$. This means $S$ consists of sequences that are zero outside a finite interval.

2. **Image of $S$ under $F$**:
- Since $F$ is injective, the image $F(S)$ has $2^N$ elements because there are $2^N$ different sequences in $S$.

3. Cutting Strings:
- Consider the strings in $F(S)$ from character $-99$ to character $N+100$. Since $F$ is injective, these strings must be distinct. Therefore, we have $2^N$ different strings of length $N+200$.

4. Probability Argument:
- The probability that a given string of length $k$ is in the image is at least $2^{-200}$. This is because the number of distinct strings of length $N+200$ in $F(S)$ is $2^N$, and the total number of such strings is $2^{N+200}$.

5. Density of Finite Strings:
- Since every finite string can be found in the image of $F$, the image of $F$ contains all finite strings.

6. Compactness and Continuity:
- The space $\{0,1\}^{\mathbb{Z}}$ with the product topology is compact and Hausdorff. The function $F$ is continuous in this topology because if two sequences are equal in a sufficiently large part, their images under $F$ will also be equal.
- The image of a continuous function from a compact space is compact. Since the image contains all finite strings and is closed, it must be dense in $\{0,1\}^{\mathbb{Z}}$.

7. Conclusion:
- Since the image is dense and closed in a compact space, it must be the entire space. Therefore, $F$ is surjective.

$\blacksquare$

### Part (b)

To determine if the statement is correct when we have a different function $f_i$ for each $i$:

1. **Different Functions $f_i$**:
- If we have a different function $f_i$ for each $i$, the mapping $F$ is no longer defined by a single function $f$ but by a sequence of functions $\{f_i\}$.

2. Injectivity and Surjectivity:
- The injectivity of $F$ does not necessarily imply surjectivity in this case. The argument used in part (a) relies on the continuity and compactness properties, which may not hold if each $f_i$ is different.

3. Counterexample:
- Consider a sequence of functions $\{f_i\}$ such that each $f_i$ only depends on $a_i$. For example, $f_i(a_i) = a_i$. This mapping is injective but not surjective because it cannot produce sequences that are not already in the original set.

Therefore, the statement is not necessarily correct if we have a different $f_i$ for each $i$.

The final answer is False