(a) The sum of the independent Poisson variables has distribution \(D+E\sim\operatorname{Po}(1.7+2.1)=\operatorname{Po}(3.8)\). Because the sum is integer-valued, \(D+E<4\) means \(D+E\le3\):

\[P(D+E<4)=e^{-3.8}\left(1+3.8+\frac{3.8^2}{2!}+\frac{3.8^3}{3!}\right)=0.4734848\ldots=\boxed{0.473}\quad\text{(3 s.f.)}.\]

(b) If \(D=1\) and \(D+E<4\), then \(E\) is 0, 1 or 2. By independence,

\[P(D=1,\ D+E<4)=P(D=1)P(E\le2)=e^{-1.7}(1.7)e^{-2.1}\left(1+2.1+\frac{2.1^2}{2!}\right)=0.2017508\ldots.\]

Hence \[P(D=1\mid D+E<4)=\frac{0.2017508\ldots}{0.4734848\ldots}=\boxed{0.426}\quad\text{(3 s.f.)}.\]

(c) The total has distribution \(T\sim\operatorname{Po}(170)\). Since the mean is large, approximate it by \(Y\sim\mathrm N(170,170)\).

The event \(T>190\) is \(T\ge191\), so the continuity-corrected boundary is \(190.5\):

\[P(T>190)\approx P(Y>190.5)=1-\Phi\left(\frac{190.5-170}{\sqrt{170}}\right)=1-\Phi(1.57228\ldots)=\boxed{0.0579}\quad\text{(3 s.f.)}.\]

Normal tables may give 0.0580; both values are accepted by the official MS.