= Solution
For the original <exponential distribution>, cancel the nonzero root in
$$
\frac{1}{1-\mu R}-1=(1+\theta)\mu R
$$
to obtain
$$
\boxed{R_{\rm old}=\frac{\theta}{(1+\theta)\mu}.}
$$
It lies strictly below the transform pole $1/\mu$.
With the extra expenses, the insurer's payment per claim is $Y=X+A$, where $A$ has <exponential distribution> of <expected value> $\mu/2$ and is independent of $X$. The <convolution of independent random variables> gives a <hypoexponential distribution> with
$$
\mathbb EY=\frac32\mu,\qquad M_Y(r)=\frac{1}{(1-\mu r)(1-\mu r/2)},\quad r<1/\mu.
$$
Keeping the <relative safety loading> fixed means using the new expected payment: the premium rate becomes $c_{\rm new}=\lambda(1+\theta)(3\mu/2)$. It does not mean keeping the old premium rate fixed. The <adjustment coefficient with independent claim expenses> therefore solves
$$
\frac{1}{(1-\mu R)(1-\mu R/2)}-1=(1+\theta)\frac32\mu R.
$$
Set $t=\mu R$, cancel $t>0$, and simplify:
$$
3(1+\theta)t^2-(7+9\theta)t+6\theta=0.
$$
The quadratic is positive at $t=0$ and equals $-4$ at $t=1$. Its leading coefficient is positive, so the smaller root is in $(0,1)$ and the larger root exceeds $1$. Only the smaller root lies in the finite <moment-generating function> domain. Hence
$$
\boxed{R_{\rm new}=\frac{7+9\theta-\sqrt{9\theta^2+54\theta+49}}
{6(1+\theta)\mu}.}
$$
For $\theta=1$,
$$
\boxed{R_{\rm old}=\frac{1}{2\mu},\qquad
R_{\rm new}=\frac{4-\sqrt7}{3\mu}\approx\frac{0.451416}{\mu},\qquad
\frac{R_{\rm new}}{R_{\rm old}}=\frac{2(4-\sqrt7)}{3}\approx0.902832.}
$$
Thus \b[the new adjustment coefficient is about $9.72\%$ smaller], even though the premium rate has been increased to retain the same <relative safety loading>. The <Lundberg inequality> consequently has a slower exponential decay rate as a function of capital.
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