(1) You do not give the $T$-statistic, which is required to find the P-value, independently of the computer output.
Furthermore, I cannot replicate your P-value from the numbers you give. Using Minitab's one-sample t test of $H_0: \mu = 130$ vs $H_a: \mu \ne 130.$
One-Sample T
Test of μ = 130 vs ≠ 130
N Mean StDev SE Mean 95% CI T P
200 140.00 25.00 1.77 (136.51, 143.49) 5.66 0.000
Minitab shows a P-value of 0.000
, which means less than $0.0005.$ A test against the one sided alternative $H_a: \mu > 130$ would be half as large---still less than $0.0005.$
(2) Usually, you cannot find a P-value from printed tables. The difficulty is not mainly that degrees of freedom (rows) jump from 100 to 1000. There is not much change in tabled values between those two rows.
The
main difficulty is not most printed t tables show too few columns, so that you can only 'bracket' the P-value between probabilities given in two adjacent column headers.
(3) A right-sided P-value corresponding to test statistic $T$ is the Probability (area) beneath the density curve of
$\mathsf{T}(\nu = 200-1 = 199)$ to the right of $T.$
One can use statistical software to find an exact P-value (very small). For $T = 5.66$ (as in the computer output above)
is as shown in the computation from R below:
1 - pt(5.66, 199)
[1] 2.613768e-08
If you had $T = 2.40,$ then the one-sided P-value would be 0.0087$ (to four places).
1-pt(2.4, 199)
[1] 0.0086583
Note: Above I said there is not much change between
degrees of freedom $\nu = 100$ and $\nu = 1000.$ For $T = 2.40,$ the corresponding P-values for sample sizes
$n = 101$ and $n = 1001$ be about $0.009$ and $0.008,$ respectively.
1-pt(2.4, 100)
[1] 0.009122769
1-pt(2.4, 1000)
[1] 0.008288541