Answer: (a) \(\frac{243}{8}\). (b) \(512\).

Let the first term be \(a\) and the common ratio be \(r\), where \(r\gt 0\).

The second and third terms are \(ar\) and \(ar^2\), so

\(ar+ar^2=168.\)

Factorising gives

\(ar(1+r)=168. \tag{1}\)

The fourth and fifth terms are \(ar^3\) and \(ar^4\), so

\(ar^3+ar^4=94.5.\)

Thus

\(ar^3(1+r)=94.5. \tag{2}\)

Divide equation \((2)\) by equation \((1)\):

\(r^2=\dfrac{94.5}{168}=\dfrac{945}{1680}=\dfrac{9}{16}.\)

Since \(r\gt 0\),

\(r=\frac34.\)

Use \((1)\):

\(a\cdot\frac34\cdot\frac74=168.\)

So

\(a\cdot\frac{21}{16}=168,\)

and hence

\(a=128.\)

(a) The sixth term is

\(ar^5=128\left(\frac34\right)^5=\frac{243}{8}.\)

(b) Since \(|r|\lt 1\), the sum to infinity exists:

\(S_\infty=\dfrac{a}{1-r}=\dfrac{128}{1-\frac34}=512.\)