= Solution
The R command fits a <Poisson regression> with the default <log link>:
$$
Y_i\sim\operatorname{Poisson}(\lambda_i),\qquad
\log\lambda_i=\beta_0+\beta_1T_i,
$$
with independent annual counts conditional on the supplied <temperature> index. The fitted mean is $\widehat\lambda(T)=\exp(2.26061+0.48870T)$. The intercept gives about $9.59$ expected named storms at index zero. A one-unit increase in the index multiplies the expected count by $e^{0.48870}\simeq1.63$; a $0.1$ increase multiplies it by about $1.050$.
The reported standard errors estimate the sampling uncertainty of the coefficients. Each displayed $z$ is the estimate divided by its <standard error>, compared with a standard <normal distribution> under a zero-coefficient null hypothesis. The slope has $z=2.879$ and two-sided $p=0.00399$, evidence of positive association within this model. The intercept test is against mean one at index zero and is not itself a test of a climatic mechanism. Association does not establish causation.
There are $61$ annual observations and two fitted coefficients, giving $59$ residual degrees of freedom. The residual <deviance> is
$$
2\sum_i\left[Y_i\log(Y_i/\widehat\lambda_i)-(Y_i-\widehat\lambda_i)\right]=51.499,
$$
with $0\log0=0$. It is not large relative to $59$ and gives no obvious overdispersion signal, although temporal dependence and omitted covariates still require substantive assessment.
For 2005 insert the index into the fitted mean:
$$
\boxed{\widehat\lambda_{2005}=e^{2.26061+0.48870(0.743)}\simeq13.79}.
$$
This is an expected count, approximately fourteen storms, rather than a guarantee of an integer outcome. In R one would use `predict(Mod, newdata=data.frame(Temp=0.743), type="response")`. A predictive distribution must include Poisson variation and, if desired, coefficient-estimation uncertainty; the latter needs the coefficient <covariance matrix>, not just the two printed standard errors.
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