= Solution
Each `scan()` command reads a numeric vector from the entered data: the first creates the nine counts, and the second their nine predictor values. The last command fits a <generalized linear model> with the <Poisson distribution> and its default <log link>, including an intercept, then prints a summary. Thus the model is <independent> counts with means $\mu_j$ satisfying
$$
\log\mu_j=b_0+b_1i_j,
\qquad \widehat\mu_j=\exp(1.363+0.3106i_j).
$$
The coefficients are <maximum-likelihood estimates>, and the reported standard errors come from the inverse fitted <Fisher information matrix>. Increasing the predictor by one multiplies the fitted mean by $e^{0.3106}\simeq1.364$, about a $36.4\%$ increase. The intercept corresponds to a mean $e^{1.363}\simeq3.91$ at predictor zero, outside the observed range. Approximate coefficient-to-standard-error ratios are $6.17$ and $8.13$; the slope is strongly different from zero under the assumed model. A $95\%$ <Wald confidence interval> for the slope is about $[0.236,0.385]$, or $[1.266,1.470]$ for the multiplicative change.
The <deviance> is twice the log-likelihood advantage of the <saturated model> over the fitted model. For these counts it is
$$
D=2\sum_{j=1}^9\left[n_j\log\frac{n_j}{\widehat\mu_j}-(n_j-\widehat\mu_j)\right],
$$
with the logarithmic term interpreted as zero when a count is zero. The <saturated model> has nine freely fitted means; the fitted model has two <independent> regression coefficients, so the residual <degrees of freedom> are \b[$9-2=7$].
Under an adequate Poisson model and the usual large-sample approximation, $D$ is compared with $\chi^2_7$. The reported $13.218$ is rather large relative to seven, but below the $5\%$ upper critical value about $14.07$; its approximate upper-tail <probability> is $0.067$. Thus there is some indication of lack of fit or <overdispersion>, but not a rejection at the $5\%$ level on this diagnostic. The deviance-to-d.f. ratio is about $1.89$; it does not by itself prove an overdispersed model, and the chi-square approximation is imperfect for small counts.
There is one inconsistency in the supplied numerical output. Independently solving the Poisson score equations for the displayed nine counts gives $\widehat b_0=1.363454$, $\widehat b_1=0.310563$ and deviance $13.217971$, agreeing with the printed estimates and deviance. But $[X^{\mathsf T}\operatorname{diag}(\widehat\mu)X]^{-1}$ gives standard errors approximately $0.221295$ and $0.030850$. In particular the printed slope standard error $0.0382$ is not the standard Poisson-model value for these data. The interpretations above describe the supplied numbers; the directly recomputed slope ratio is $10.07$ and its normal-approximation $95\%$ interval is about $[0.2501,0.3710]$. The conclusion of a strong increasing trend is unchanged.
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