Certainly $a_1 \le 3 \lt 4$, since $a_1=3$.
Suppose that for the specific integer $k$, we know that $3\le a_k \lt 4$.
We want to show that similar inequalities hold for the "next" term $a_{k+1}$. That is, we want to show that $3\le a_{k+1}\lt 4$.
There are two inequalities to prove. Let us deal with them separately. First we show that (i) $3\le a_{k+1}$ and then that (ii) $a_{k+1}\lt 4$.
(i) Note that $a_{k+1}=4-\frac{2}{a_k}$. We know that $a_k \ge 3$. So $\frac{2}{a_k}\le \frac{2}{3}$. It follows that $4-\frac{2}{a_k} \ge 4-\frac{2}{3}=\frac{10}{3} \ge 3$.
(ii) Because $a_k$ is positive, $4-\frac{2}{a_k}\lt 4$.
So we have shown that our inequalities hold at $n=1$, and that if they hold for some integer $k$, they hold for the next integer $k+1$. By the principle of mathematical induction, the inequality $3\le a_n \lt 4$ holds for every positive integer $n$.
Added: Since the above answer was written, the question has quintupled in length. I hope that the above will at least help with the mechanics of induction. We make a few comments about the added parts.
The added question (b) is undoubtedly something you can handle. Showing that $3\lt \alpha \lt 4$ can be done by observing that $1\lt \sqrt{2}\lt 2$. To show that $\alpha=4-\frac{2}{\alpha}$, note that $\frac{2}{\alpha}=\frac{2}{2+\sqrt{2}}$. Multiply top and bottom by $2-\sqrt{2}$, and everything will collapse. Alternately, we want to show that $\alpha^2=4\alpha -2$. Solve the quadratic equation $x^2-4x+2=0$ using the Quadratic Formula. You will find that the roots are $2\pm\sqrt{2}$.
For (c), the natural thing is to take the left-hand side $a_{n+1}-\alpha$, and replace $a_{n+1}$ by its value in terms of $a_n$. Also, we use the result of (b) as a hint, and replace $\alpha$ by $4-\frac{2}{\alpha}$. We get $a_{n+1}-\alpha=\left(4-\frac{2}{a_n}\right)-\left(4-\frac{2}{\alpha}\right).$ There is some nice cancellation: The expression on the right simplifies to $-\frac{2}{a_n}+\frac{2}{\alpha}$. Bring to a common denominator and we get $a_{n+1}-\alpha=\frac{2(a_n-\alpha)}{a_n \alpha}.$
Now you be able to show (d). For the above formula gives us information about how close $a_{n+1}$ is to $\alpha$ in terms of how close $a_n$ is to $\alpha$. Use the fact that since $a_n \gt 3$ and $\alpha\gt 3$, we have $\frac{2}{a_n \alpha}\lt \frac{2}{9}\lt \frac{1}{4}$.