How to open a particular link with selenium?

I would use

elem = driver.find_element_by_class_name("jobs-u-font--bold")

To get the <span>, since that seems like a unique class name (although I can't be sure from your post). Then you can reach the <a> level with

a_elem = elem.find_element_by_xpath("..")

Then you can a_elem.click() or whatever you are trying to do.

To access the anchor tag you can use css selector with attribute data-tn-element="view-unread-candidates" i believe it should be same for all anchor tag.

elements=driver.find_elements_by_css_selector('a[data-tn-element="view-unread-candidates"]')
for ele in elements:
    print(ele.get_attribute("href"))

Or if you want to use child element and then want to fetch the parent tag then try below code with xpath.

elements=driver.find_elements_by_xpath("//span[@class='jobs-u-font--bold']")
for ele in elements:
    print(ele.find_element_by_xpath("./parent::a").get_attribute('href'))

I would use:

List elements = driver.findElements(By.xpath("//a[@data-tn-element='view-unread-candidates']"));

    Iterator<WebElement> iter = elements.iterator();

    while (iter.hasNext()) {
        WebElement item = iter.next();
        String href = item.getAttribute("href");
        System.out.println("href is " + href);
    }
}

And if you want to click the link with the particular href, then you can put the if condition after getting the href in the above code . When that condition will meet, click on the element.