Answer
(a) $\color{blue}{\approx 0.0053}$
(b) $\color{blue}{\approx 0.0197}$
Work Step by Step
(a)
$\begin{align*}
P(X=200) &= {400 \choose 200}(0.55)^{200}(.45)^{200} \\
&\approx (1.0295\times 10^{119})(1.18178\times 10^{-52})(4.39039\times 10^{-70}) \\
P(X=200)\ &\color{blue}{\approx 0.0053}
\end{align*}$
(b)
A normal approximation may be used since $n=400 \gt 9\left(\dfrac{p}{q}\right) = 9\left(\dfrac{.55}{.45}\right) = 11$ and $n=440 \gt 9\left(\dfrac{q}{p}\right) = 9 \left(\dfrac{.45}{.55}\right) \ge 8$.
Since $X\sim \text{Binomial}(n=400,p=.55)$, then $X\stackrel{\Large\cdot}{\sim} N(\mu = np = 400(.55), \sigma^2 = npq = 400(.55)(.45))$.
$\begin{align*}
P(X\le 199) &= \sum_{k=0}^{199} {400 \choose k}(0.55)^k(.45)^{400-k} \\
&= P\left( \dfrac{X-\mu}{\sigma} \le \dfrac{199 -400(.55)}{\sqrt{400(.55)(.45)}}\right) \\
&= P\left( \dfrac{X-\mu}{\sigma} \le \dfrac{199 -400(.55)+0.5}{\sqrt{400(.55)(.45)}}\right),\quad \text{[ continuity correction ]} \\
&\approx P\left( Z \le \dfrac{-20.5}{\sqrt{99}} \right),\ Z\sim N(0,1) \\
&\approx P(Z\le -2.06 ) \\
&= F_Z(-2.06) \\
P(X\le 199)\ &\color{blue}{\approx 0.0197} \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]}
\end{align*}$
