This kind of problem has been solved quite a few times on this site, but I want to try to show you how you could simply have followed the road map laid out in the statement of the problem.
In (a) you should at least try to follow the hint. Clearly $a_1^2=4\ge 2$, so you can immediately go on to the induction step: assume that $a_n^2\ge 2$, and try to use that to show that $a_{n+1}^2\ge 2$. There is only one reasonable way to approach this: start by expressing $a_{n+1}^2$ in terms of $a_n$.
$$\begin{align*}
a_{n+1}^2&=\frac14\left(a_n+\frac2{a_n}\right)^2\\
&=\frac14\left(a_n^2+4+\frac4{a_n^2}\right)\\
&=\frac{a_n^2}4+1+\frac1{a_n^2}\,.
\end{align*}$$
At this point you could try applying the induction hypothesis in a very straightforward way:
$$\begin{align*}
\frac{a_n^2}4+1+\frac1{a_n^2}&\ge\frac24+1+\frac1{a_n^2}\\
&=\frac32+\frac1{a_n^2}
\end{align*}$$
Unfortunately, $\frac1{a_n^2}<\frac12$, so that doesn’t yield the desired result. We need to dig a little deeper. We’re assuming that $a_n^2\ge 0$, and we want to know that $a_{n+1}^2-2\ge 0$, so perhaps we should focus more narrowly on $a_n^2-2$ and $a_{n+1}^2-2$. Let $d=a_n^2-2$; then
$$\begin{align*}
a_{n+1}^2-2&=\frac{a_n^2}4+1+\frac1{a_n^2}-2\\
&=\frac{2+d}4-1+\frac1{2+d}\\
&=\frac{(2+d)^2-4(2+d)+4}{4(2+d)}\\
&=\frac{d^2}{4(2+d)}\\
&\ge 0\,,
\end{align*}$$
as desired.
In (b) there is also a very reasonable place to begin, even before you have any real idea of how to proceed. Clearly we want to look at $a_n-a_{n+1}$, and the equation defining the relationship between $a_{n+1}$ and $a_n$ is the obvious place to start: we’re given that $a_{n+1}=\frac12\left(a_n+\frac2{a_n}\right)$, we have $2a_{n+1}=a_n+\frac1{a_n}$, and therefore
$$a_n-a_{n+1}=a_{n+1}-\frac1{a_n}\,.$$
The hint indicates that (a) is useful, so what does it say about $a_n$ and $a_{n+1}$? It says that $a_n^2\ge 2$ and $a_{n+1}^2\ge 2$, and since it’s clear (and easily proved by induction on $k$) that all of the numbers $a_k$ are positive, this means that $a_{n+1}\ge\sqrt2$ and $a_n\ge\sqrt2$. Finally, the fact that $a_n\ge\sqrt2$ implies that
$$\frac1{a_n}\le\frac1{\sqrt2}<\frac2{\sqrt2}=\sqrt2\,,$$
and we conclude that $a_n-a_{n+1}>0$ for all $n$.
Now (c) should be immediate: the sequence $\langle a_n:n\ge 1\rangle$ is monotone decreasing and bounded below, so … ?
(d) uses a standard trick that you can employ whenever you know that a sequence actually has a limit: if $\lim\limits_{n\to\infty}a_n=L$, then $\lim\limits_{n\to\infty}a_{n+1}=L$ as well. Thus,
$$\begin{align*}
L&=\lim_{n\to\infty}a_{n+1}\\
&=\lim_{n\to\infty}\frac12\left(a_n+\frac2{a_n}\right)\\
&=\frac12\left(\lim_{n\to\infty}a_n+\frac2{\lim\limits_{n\to\infty}a_n}\right)\\
&=\frac12\left(L+\frac2L\right)\,,
\end{align*}$$
and you can now solve for $L$.
Finally, (e) just asks you to go back through the work done in (a)-(d) and see what has to be modified to make the sequence converge to $\sqrt{c}$ instead.
