If you don't care about the limit function being continuous, one of the simplest examples is the sequence of functions $f_{n}(x) = x^{n}$ on $[0,1]$. This sequence converges pointwise to the function which is zero for $x \lt 1$ and $1$ for $x = 1$. This is of course an example on $[0,1]$, but you get an example on $\mathbb{R}$ by extending all the functions by zero for $x \leq 0$ and by $1$ for $x \geq 1$.
Recall that a continuous function on a compact interval is automatically uniformly continuous (prove this, in case you don't know that statement!). To get an example of a non-uniformly continuous function, we need to look for a function on an unbounded interval.
A very simple example of a continuous but not uniformly continuous function is $f(x) = x^{k}$ on $[0,\infty)$ for $k \neq 0,1$. Now simply define $f_{n}(x) = f(x)$ if $0 \leq x \leq n$ and $f_{n}(x) = f(n)$ if $x \geq n$. You should be able to check yourself that each $f_{n}$ is uniformly continuous and that the sequence $f_{n}$ converges pointwise to $f$.
To extend this example to all of $\mathbb{R}$, simply extend the functions by zero to the left.
A very similar idea works for $e^{x}$. If you want a slightly more interesting example, you can try to tackle $f(x) = \sin{(e^x)}$ with the same idea of truncating and extending constantly to the left and right.
Added: In fact, the procedure I outlined is one that always works (there are other ways but this probably is the most straightforward one). More precisely, if $f: \mathbb{R} \to \mathbb{R}$ is continuous, put
$$f_{n}(x) = \begin{cases} f(x), &\text{if } |x| \leq n,\\f(n), &\text{if }x \geq n,\\f(-n), &\text{if } x \leq -n\end{cases}$$
and check that $f_{n}$ is uniformly continuous. This is because $f_{n}$ is uniformly continuous inside the compact interval $[-(n+1),n+1]$ due to the fact I mentioned above and constant outside. It is easy to see that $f_{n}(x) \to f(x)$ for all $x$, so $f_{n} \to f$ pointwise.
So to get an example of the kind you're asking about, the only thing you really need to think about is how to find a continuous but not uniformly continuous function, and I've given a few examples that should illustrate the kinds of functions you should be looking at.
Using Alex Ravsky's hint, one such counterexample for part (b) is $f_n(x)=x-\frac{1}{n}$ for each $n$, and $g(x)=x^2$. Here is my full solution:
First of all, $f_n(x)$ converges uniformly to $f(x)=x$, since $$\sup_{x\in R} |f_n(x)-f(x)|=\frac{1}{n}\to 0 \ \ \ \ \ \textrm{as } n\to\infty$$
We claim that the sequence of functions defined by $h_n(x)=g(f_n(x))=(x-\frac{1}{n})^2$ converges pointwise, but not uniformly to $h(x)=x^2$. Indeed, for given $\epsilon>0$ and $x\in\mathbb{R}$, we can choose $N$ so large that
$$\frac{|2x|}{N}+\frac{1}{N^2}<\epsilon$$
Then for any $n\ge N$, we get
$$ |h_n(x)-h(x)|=\left|x^2-\frac{2x}{n}+\frac{1}{n^2} - x^2\right|\le \frac{|2x|}{n}+\frac{1}{n^2}\le\frac{|2x|}{N}+\frac{1}{N^2}<\epsilon$$
which proves pointwise convergence. The convergence is not uniform because
$$|h_n(n)-h(n)|=\left|\left(n-\frac{1}{n}\right)^2-n^2\right|=2-\frac{1}{n^2}\ge 1$$
for each $n\in\mathbb{N}$. In particular,
$$\sup_{x\in R} |h_n(x)-h(x)|\not\to 0\ \ \ \ \ \textrm{as } n\to\infty$$
Best Answer
Yes, take
$$f_n(x)=\left\lbrace\begin{array}{cc}0&x\neq 0\\\frac{1}{n} & x=0\end{array} \right.$$